Complex Analysis Mathematics 113. Analysis I: Complex Function Theory
Complex Analysis Mathematics 113. Analysis I: Complex Function Theory
For the students of Math 113
Harvard University, Spring 2013
FOREWORD
Welcome to Mathematics 113: Complex Function Theory! These are the compiled lecture notes for the class, taught by Professor Andrew Cotton-Clay at Harvard University during the spring of 2013. The course covers an introductory undergraduate-level sequence in complex analysis, starting from basics notions and working up to such results as the Riemann mapping theorem or the prime number theorem. We hope that these lecture notes will be useful to you in the future, either as memories of the class or as a handy reference. These notes may not be accurate, and should not replace lecture attendance.
Instructor: Andy Cotton-Clay,
[email protected] Course assistants: Felix Wong,
[email protected], Anirudha Balasubramanian,
[email protected] Course websites: Professor Cotton-Clay’s website, http://math.harvard.edu/~acotton/ math113.html, Harvard course website http://courses.fas.harvard.edu/0405
2
Cover image: domain colored plot of the meromorphic function f (z) = (z−1)(z+1) . (z+i)(z−i)2 Source: K. Poelke and K. Polthier. Lifted Domain Coloring. Eurographics/ IEEE-VGTC Symposium on Visualization, (2009) 28:3. Credits: V.Y. Downloaded from www.flickr.com/photos/syntopia/6811008466/sizes/o/in/photostream/.
1
TABLE OF CONTENTS
The contents of this compilation are arranged by lecture, and generally have the topics covered in a linear manner. Additional notes are provided in the appendices.
Foreword
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
Introduction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2
Riemann Sphere, Complex-Differentiability, and Convergence . . . . .
7
3
Power Series and Cauchy-Riemann Equations . . . . . . . . . . . . . . . 11
4
Defining Your Favorite Functions . . . . . . . . . . . . . . . . . . . . . . . 15
5
The Closed Curve Theorem and Cauchy’s Integral Formula . . . . . . 18
6
Applications of Cauchy’s Integral Formula, Liouville Theorem, Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
7
Mean Value Theorem and Maximum Modulus Principle . . . . . . . . 28
8
Generalized Closed Curve Theorem and Morera’s Theorem . . . . . . 31
9
Morera’s Theorem, Singularities, and Laurent Expansions . . . . . . . 34
10
Meromorphic Functions and Residues . . . . . . . . . . . . . . . . . . . . 40
11
Winding Numbers and Cauchy’s Integral Theorem . . . . . . . . . . . . 42
12
The Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
13
Integrals and Some Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 49
14
Fourier Transform and Schwarz Reflection Principle . . . . . . . . . . . 52
2
15
M¨ obius Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
16
Automorphisms of D and H . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
17
Schwarz-Christoffel and Infinite Products . . . . . . . . . . . . . . . . . . 62
18
Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 65
19
Riemann Mapping Theorem and Infinite Products . . . . . . . . . . . . 69
20
Analytic Continuation of Gamma and Zeta . . . . . . . . . . . . . . . . . 72
21
Zeta function and Prime Number Theorem . . . . . . . . . . . . . . . . . 78
22
Prime Number Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
23
Elliptic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
24
Weierstrass’s Elliptic Function and an Overview of Elliptic Invariants and Moduli Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
A
Some Things to Remember for the Midterm . . . . . . . . . . . . . . . . 94
B
On the Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
C
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Afterword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3
CHAPTER
1 INTRODUCTION
√ Consider R but throw in i = −1 to get C = {a + bi : a, b ∈ R}. We will be looking at functions f : C → C which are complex-differentiable. Complex-differentiable, or holomorphic, functions are quite a bit different from real-differentiable functions. We can think of the real world as rigid, and the complex world as flexible. Complex analysis has applications in topology and geometry (in particular, consider complex 4-manifolds), and of course, physics and everywhere else. Anyway, write C = R2 with the identification (a, b) = a + bi. Addition and multiplication by a ∈ R is as for R. For multiplication in general, (a, b)(c, d) := (ac − bd, ad + bc). Claim. C is a field: in particular, it’s an additively commutative group, and multiplicatively, C − {0} is a commutative group. Proof. Should be immediate. Claim. Every nonzero complex number z ∈ C has a multiplicative inverse. b a Proof. In fact, z −1 := a2 +b 2 − ( a2 +b2 )i works.
For z ∈ C, z = a + bi, we define the complex conjugate z := a − bi = a + b(−i). Claim. z + w = z + w, zw = zw for z, w ∈ C. Proof. Write it out. We also define the norm |z| := norm:
√
a2 + b2 = zz. We can deduce a few properties of the
Claim. (multiplicative property) |z||w| = |zw| for z, w ∈ C. Proof. It suffices to work with squares. |z|2 |w|2 = zzww = zwzw = |zw|2 .
√ Algebraically, can we do better than the complex numbers (e.g. can we throw in 1 + i, etc.)? The Fundamental Theorem of Algebra tells us we’re done as soon as we throw in i;
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that is, we get all the solutions to polynomials. P In particular, let p : C → C, p(z) = ak z k for ai ∈ C. If the degree is > 0, then ∃z ∈ C : p(z) = 0. Consider the equation z 2 − (1 + i) = 0. Let’s work this out explicitly for square roots. Given a + bi, we want x + iy s.t. (x + iy)2 = a + bi, which gives a system of equations: ( ( x2 − y 2 = a a2 = (x2 − y 2 )2 = x4 − 2x2 y 2 + y 4 =⇒ =⇒ (x2 −y 2 )2 +(2xy)2 = a2 +b2 2xy = b (x2 + y 2 )2 = x4 + 2x2 y 2 + y 4 ( √ x2 = 21 (a + a2 + b2 ) ≥ 0 √ y 2 = 12 (−a + a2 + b2 ) ≥ 0 ! p p √ √ a + a 2 + b2 −a + a2 + b2 + isgn(b) 2 2
p =⇒ x + y = a2 + b2 =⇒ 2
2
=⇒ x + iy = ±
We can take this big general formula and use it to solve our specific cases. For z ∈ C, z = a + bi, we define the real part Re(z) = a = Im(z) = b = z−z 2i .
z+z 2 ,
the imaginary part
The Geometry of the Complex Plane. We have a multiplicative norm |z|. Viewing z, w ∈ C as vectors, addition is visualized as a parallelogram, and multiplication is best seen in polar coordinates. Polar coordinates can be obtained by r = |z| and θ = Arg(z), the argument of z. So we get the pretty important identity z = r cos θ + ir sin θ = r(cos θ + i sin θ). For multiplication, try z1 z2 , where z1 = r1 (cos θ1 + i sin θ1 ), z2 = r2 (cos θ2 + i sin θ2 ). Then z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )). So |z1 z2 | = |z1 ||z2 | and Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ) mod 2π. Namely, multiplication by z = r(cos θ + i sin θ) dilates the complex plane by r and rotates it counterclockwise by θ. Another view of the complex number is that z = r(cos θ + i sin θ) acts on C ∼ = R2 by cos θ − sin θ r cos θ −r sin θ a −b r = = . sin θ cos θ r sin θ r cos θ b a Of course, i sin(nθ)).
1 z
=
1 r (cos(−θ)
+ i sin(−θ)) =
1 r (cos θ
− i sin θ) =
1 |z|2 z.
So z n = rn (cos(nθ) +
Problem. Find all cube roots of 1. Solution. Want z 3 = 1 = 1(cos 0 + i sin 0). So r3 = 1 =⇒ r = 1 and 3θ = 0 mod 2π =⇒ 4π θ = 0, 2π 3 , 3 . These give the cube roots of unity, and we can plot them on the unit circle in C. 2π th Notation. ζn = cos 2π root of unity. The nth roots of unity n + i sin n , a primitive n 2πk are ζnk = cos 2πk n + i sin n , k = 0, ..., n − 1.
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The equation for nth roots of unity is z n − 1 = 0 = (z − 1)(z n−1 + z n−2 + ... + z + 1). This gives a simple formula for 1 + cos θ + cos 2θ + ... + cos(n − 1)θ. For z = cos θ + i sin θ, P Pn−1 1 + z + ... + z n−1 = ( k=0 cos kθ + i sin kθ). Now multiply by z − 1 = (cos θ − 1) + i(sin θ) to get (cos nθ − 1) + i sin nθ (cos nθ − 1)(cos θ − 1) − sin nθ sin θ . = (cos θ − 1) + i sin θ (cos θ − 1)2 + sin2 θ So
n−1 X k=0
cos kθ =
(cos nθ − 1)(cos θ − 1) − i sin nθ sin θ . 2 − 2 cos θ
This isn’t the best formula, but we’re happy with it.
6
CHAPTER
2
RIEMANN SPHERE, COMPLEX-DIFFERENTIABILITY, AND CONVERGENCE
Stereographic projection. Consider a sphere S 2 := {(u, v, w) : u2 + v 2 + w2 = 1} in R3 . We have a bijection ϕ : S 2 − N → C, where N is the north pole (point at ∞), given by ˆ := C ∪ {∞} ∼ stereographic projection. Indeed, we define the Riemann sphere C = S2.
Diagram 1: Stereographic projection. Source: Wikipedia Given (u, v, w) ∈ S 2 − N and (x, y) ∈ R2 , the bijection is given by v 2x 2y x2 + y 2 − 1 u , , ϕ−1 : (x, y) 7→ , , . ϕ : (u, v, w) 7→ 1−w 1−w x2 + y 2 + 1 x2 + y 2 + 1 x2 + y 2 + 1 Claim. Any circle on S 2 maps to a circle or line in C under ϕ and vice-versa. Proof. I hope you paid attention in lecture.
Now consider the map (u, v, w) 7→ (u, −v, −w), which swaps 0 ∈ C with S, the south ˆ with N , the north pole. What does it do to z = x + iy? We see that pole, and ∞ ∈ C u−iv z 7→ 1+w .
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Claim. This is z1 . Proof. Trivial. In fact,
1 z
is just rotating the sphere around.
Claim. The inversion
1 z
sends (circles and lines) to (circle and lines).
If we have a function f : C → C and we think it’s “nice” at ∞, how do we quantify that? We simply use the inversion z1 . In particular, we look at f ( z1 ). Suppose f (∞) = ∞; then f (11 ) sends 0 to 0. z For example, consider a polynomial of degree n, f (z) = c0 + ... + cn z n and cn 6= 0. Take 1 zn 1 = = , c0 + ... + cn z −n c0 z n + ... + cn f ( z1 ) which is well-defined in a neighborhood of 0 ∈ C and clearly, sends 0 7→ 0. Differentiation. Lets talk about something we all like, differentiation. :-) Definition. Given f (z) : C ⊃open U → C, z ∈ U , we say f is complex-differentiable (or holomorphic) at z if f (z + h) − f (z) f 0 (z) := lim h→0 h exists for h ∈ C. Proposition. If f, g are holomorphic, then f g and f + g is as well, with (f g)0 = f 0 g + f g 0 and (f + g)0 = f 0 + g 0 . Proof. Also trivial. Example. Consider complex polynomials in z, which are holomorphic: f (z) = Pn−1 Then f 0 (z) exists and f 0 (z) = (k + 1)ck+1 z k .
P
ck z k .
Example. Let f (z) = z. By showing that the limits in the real and imaginary directions do not agree, show that this is not holomorphic. Before starting anything else, we would like to review real-differentiable functions. There are “different severities” of being real differentiable. Let f : R ⊃open U → R: 1. 2. 3. 4. 5.
real differentiable: f 0 (x) exists ∀x in domain C 1 : f 0 (x) exists and is continuous C k : f, f 0 , ..., f (k) exists and are continuous C ∞ : (smooth) all derivatives exists and are continuous P∞ real-analytic: f (x) = ck xk is a convergent power series
The punchline is that homolorphic functions are always C ∞ and analytic! (This is something that should really be appreciated.) Indeed, we oftentimes interchange holomorphic, C ∞ , and analytic in the complex world.
8
To show this, we will go back to convergence. Consider a sequence of functions {fk (z)} defined on E ⊂ C that is containable in a compact set. We say that it is pointwise convergence if limk→∞ fk (z) exists ∀z; the pointwise limit is defined as f (z). Suppose fk (z) is continuous. It’s uniformly convergent if ∀, ∃N : ∀n ≥ N : |fn (z) − f( z)| ≤ , ∀z ∈ E. Lemma. If {fk (z)} are continuous and converge uniformly on E, then pointwise limit f (z) is continuous. Proof. Use triangle inequality and a standard epsilon pushing argument. Power series. Consider f (z) =
P∞ 0
ck z k . We need a condition on |ck | for this to converge.
Definition. Given a sequence {ak }∞ 1 , we define the limit supremum lim sup ak = lim sup ak = lim ak k→∞
n→∞
k≥n
k→∞
The polynomial condition is then to look at lim sup |ck |1/k . Theorem. If lim|ck |1/k = L for L = 0, 0 < L < ∞, or L = ∞, then: P 1. If L = 0, then f (z) = ck z k converges for all z. P 2. If 0 < L < ∞, then ck z k converges for |z| < R = L1 , diverges for |z| > R, and R is called the radius of convergence. (We do not know for |z| = R.) P 3. If L = ∞, ck z k diverges ∀z 6= 0. Proof. L = 0: ∀ > 0, ∃N : ∀k > N, |ck |1/k < . Let = z1 |z|. Then |ck |1/k < 12 |z|, P k ck z k converges. We can do Re|ck | < 21k 1|z|k . We want |ck |1/k |z| < 21 , so |ck ||z|k ≤ 12 so this with the M -test, by considering partial sums, etc. 0 < L < ∞: lim|ck |1/k = L. Let |z| = R(1 − 2δ), then lim|z||ck |1/k = 1 − 2δ for R = L1 . P So ∃N : |z||ck |1/k < 1 − δ ∀n > N , and ck z k converges because |ck |z k < (1 − δ)k . Likewise, if we suppose |z| = R(1 + 2δ). Then lim|ck |1/k |z| = 1 + 2δ. So for infinitely many k, |ck |1/k |z| > 1+δ and |ck z k | > 1. The last condition is proved in the same manner. Example. Example. lim(k + 1)
P∞ 0
zk =
P∞ 0
1/k
1 1−z .
Deduce this from the theorem above.
(k + 1)z k =
1 (1−z)2 .
We see this by taking ck = 1 and showing that
= 1.
P∞ Theorem. Suppose f (z) = 0 ck z k converges for |z| < R. Then: P∞ k−1 1. converges for |z| < R. 1 kck z P∞ 0 2. f (z) exists and equals 1 kck z k−1 . Proof. (1) It suffices to check lim|ck |1/k ≤ R1 =⇒ lim(k + 1)1/l |ck |1/k ≤ lim(k + 1)1/k |ck |1/k = lim(k + 1)1/k lim|ck |1/k . (2) R = ∞: subtract and show that limits → 0.
1 R.
X f (z + h) − f (z) X 1X − kck z k−1 = ck [(z + h)k − z k ] − kck z k−1 h h 9
Note that
=
X
1 1 ck [ (z + h)k − z k − kz k−1 ]. h h
Recall the binomial theorem, (z + h)k =
k l k−l hz , l
which we can use to rewrite the expression above as =
∞ X 0
k ∞ X k X X k k−l l−1 k k−l l−1 k−1 ck [ z h − kz ]= z h l l 0 2 l=1
∞ X k X k k−l l−2 =h z h . l 0 2
We also have
X k k−l l−2 X k z h ≤ |z|k−1 = (|z| + 1)k , l l P P so ck (|z| + 1)k converges because R = ∞ for ck z k . R < ∞: Let |z| = R − 2δ, |h| < δ; then |z + h| < R − δ. Then ∞ k X X k k−l l−2 f (z + h) − f (z) X − kck z k−1 = h ck z h . h l 0 2
Write
k k(k − 1)...(k − l + 1) = . l l! k For l ≥ 2 we can bound kl ≤ k 2 l−2 . So X k k2 k−l l−2 z h ≤ 2 (|z| + |h|)k |z| l 2 and the equation above translates into =
h X 2 k ck (|z| + |h|)k , |z|2
which converges and → 0 because of the h in the front.
10
CHAPTER
3
POWER SERIES AND CAUCHY-RIEMANN EQUATIONS
P∞ Power series. Last time we were dealing with power series, f (z) = k=0 ck z k . We defined 1 the radius of convergence R = L , where L = lim|ck |1/k and either L = 0, 0 < L < ∞, or L = ∞. Inside the radius of convergence, f (z) is a convergent power series with some R. What are the consequences of this? Corollary. Inside its radius of convergence, a power series is ∞-differentiable, with expected derivatives as convergent power series. (k)
Corollary. If R > 0, then ck = f k!(0) . Proof. Take f (k) (x). The corollary follows.
There are also several uniqueness properties: P Lemma. If f (z) = ck z k is a convergent power series and f (zn ) = 0 for a sequence ∞ {zn }n=1 with zn → 0, zn 6= 0, then ck = 0 ∀k and Re(f (z)) ≡ 0. 2 Proof. c0 = f (0) = limn→∞ f (zn ) = 0. Now form g1 (z) = f (z) z = c1 + c2 z + c3 z + ... (exercise: show that this holds for some radius of convergence). Check c1 = g1 (0) = limn→∞ g1 (zn ) = limn→∞ f (z) z = 0, and induct on gi . P P Proposition. If f (z) = ak z k , g(z) = bk z k and these agree on some set accumulating at 0, then ak =Pbk ∀k, i.e. f (z) = g(z). Proof. Consider (ak − bk )z k and apply the lemma. Show that lim|ak |1/k ≤ L and lim|bk |1/k ≤ L =⇒ lim|ak − bk |1/k ≤ L. P Note: To center the power series at w ∈ C, consider ck (z − w)k , which shifts the center of the power series from 0 ∈ C to w and maintains the radius of convergence at R. Complex-differentiability. We refer to complex-differentiability as either holomorphic or complex-analytic (some texts, e.g. Ahlfors, simply like to use analytic). This is a very nice property of functions that we will be exploring in the upcoming weeks.
11
Definition. f : C → C is holomorphic at z ∈ C if f (z + h) − f (z) h
lim
h→0
exists for h ∈ C. Equivalently, ∃ a number f 0 (z) : 0 = lim
h→0
f (z + h) − f (z) − f 0 (z). h
The Cauchy-Riemann equations. By considering real and imaginary parts of holomorphic functions, we get the celebrated Cauchy-Riemann equations. Proposition. If f (z) : C → C is holomorphic at z (alternatively, by R2 ∼ = C, we can ∂f and exist and regard f (x, y) : R2 → R2 ), then ∂f ∂x ∂y i
∂f ∂f = . ∂x ∂y
(3.1)
Equivalently, for the decomposition f (x, y) = u(x, y) + iv(x, y) and ∂u ∂v ∂f = +i ∂x ∂x ∂x ∂f ∂u ∂v = +i , ∂y ∂y ∂y what we mean is that, by comparing real and imaginary parts of the equation ∂u ∂v ∂u ∂v i +i = +i ∂x ∂x ∂y ∂y, we obtain the equalities (
∂u ∂v ∂x = ∂y ∂v − ∂x = ∂u ∂y .
(3.2)
Proof. This is a bit tedious to type up, so I hope you paid attention in lecture. The proof follows from a simple computation of the partials; see any textbook. The CR conditions (1) or (2), along with the condition that the partials are continuous, ascertains that f is itself holomorphic. If the partials are not continuous, consider f (x, y) = xy(x+iy) x2 +y 2 , z 6= 0, and 0 for z = 0. Show that this is not differentiable at 0 but
∂f ∂x (0, 0)
=
∂f ∂y (0, 0)
= 0.
∂f Proposition. If f (x, y) = u(x, y)+iv(x, y) has continuous partial derivatives and i ∂f ∂x = ∂y (satisfies CR equations), then f is holomorphic. Proof. Messy as well, but use the mean value theorem and write out the partials.
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An equivalent (geometric) formulation of the CR equations. Consider the Jacobian of f : R2 → R2 : ! ∂u ∂x ∂v ∂x
Jf := and the rotation matrices
∂u ∂y ∂v ∂y
− sin θ cos θ
cos θ sin θ
.
So multiplication by i is multiplication by the matrix (aside: see the relation to complex structure) 0 −1 . 1 0 The equivalent form is that Jf I = IJf ; e.g. the Jacobian matrix commutes with rotation by π/2. Check this: ! ! ! ! ∂u ∂v ∂v ∂u ∂u ∂u ∂u − ∂u − ∂x − ∂y 0 −1 0 −1 ∂x ∂y ∂y ∂x ∂x ∂y = = = . ∂v ∂v ∂v ∂v ∂u ∂v ∂v 1 0 1 0 − ∂x − ∂u ∂x ∂y ∂y ∂x ∂y ∂x ∂y An algebraic interpretation. Consider complex polynomials, p(z) = at complex-valued polynomials of real variables x, y: n X
p(x, y) =
X
Pn
k=0 ck z
ck,l xk y l
k
. Look
(3.3)
m=0 k+l=m, k,l≥0
(0, 0), and take a “new basis” as z = x + iy and z = x − iy. Write the with ck,l = ∂ k∂p x∂ l y polynomial in z, z : X X c˜k,l z k z l . (3.4) p(z, z) = m=0 k+l=m, k,l≥0
We claim that there’s a 1-1 correspondence between polynomials given by (3) and (4). ∂ ∂ Indeed, ∂x , ∂y act on these polynomials: 1 ∂ = ∂z 2
∂ ∂ −i ∂x ∂y
∂ 1 = ∂z 2
,
∂ ∂ +i ∂x ∂y
Claim. We have the equalities ∂ k l (z z ) = lz k z l−1 , ∂z ∂ k l (z z ) = kz k−1 z l . ∂z Proof. We simply check: ∂ k l 1 i (z z ) = [kz k−1 z l + lz k z l−1 ] + [kiz k−1 z l + (−i)lz k z l−1 ] = lz k z l−1 , ∂z 2 2 13
and likewise for the second equality.
Claim. (equivalent form of CR equations) f (z) satisfies CR equations ⇐⇒ Proof. A simple check.
∂ ∂z f
= 0.
PP The conclusion is that p(z, z) = P P c˜k,l z k z l is holomorphic iff c˜k,l = 0 for l > 0 (e.g. 0 no z s).P Alternatively, p(x, y) = ck,l xk y l is holomorphic iff it can be written as k p(z) = ck z . Let’s give some basic hints that holomorphic functions behave a bit like power series: Lemma. Suppose f (x, y) = u(x, y) + iv(x, y) is holomorphic on a disk of radius R and u(x, y) is constant. Then f is constant. Proof. ux = uy = 0, so by CR equations, ux = vy , uy = −vx and vx = vy = 0. By the mean value theorem, this shows that v is constant along horizontal and vertical lines in the plane. So v is constant throughout the disk. Lemma. If f = u + iv is holomorphic on a disk in C and ||f ||2 is constant, then f is constant. Proof. We have u2 + v 2 = c, so taking partials by x and y gives 2ux u + 2vx v = 0 and 2uy u + 2vy v = 0. Applying the CR equations, we get 2ux v − 2vx u = 0. Adding equations gives 2ux (u2 + v 2 ) = 0, and similarly 2vx (u2 + v 2 ) = 0. So either u2 + v 2 = 0 (f = 0), or all partials = 0. In the second case, given any open ball, f is constant in that open ball since its partials = 0 in the ball. But the disk is connected, so f is constant everywhere.
14
CHAPTER
4 DEFINING YOUR FAVORITE FUNCTIONS
Today is about extending your favorite (real analytic) functions from R to C. When determining their complex analogues, we could (1) ask for the same fundamental properties; (2) use their power series; (2’) extend such that the result is holomorphic (actually, this is the same as 2). Happily, these all agree! 1. ex d x e = ex , ... or (2) ex = We can consider (1) e0 = 1, dx
P xn n!
.
x
(1) Let’s look for f (z) with f (x) = e and f (z1 + z2 ) = f (z1 )f (z2 ) and holomorphic. Apparently, f (x + iy) = ex f (iy). Let f (iy) = A(y) + iB(y), so f (x + iy) = ex A(y) + ex iB(y). Then use the Cauchy-Riemann equations to show that A(y) = cos y and B(y) = sin y by the Fundamental Theorem of ODEs (namely, any linear ordinary differential equation has an ez solution). So define ez = ex (cos y + i sin y), for z = x + iy. (2) Let f (z) =
∞ X zn . n! n=0
1 1/n The radius of convergence is ∞, since limn→∞ ( n! ) = 0 (e.g. it’s an entire function). Also, f (z) is holomorphic on all of C. We can easily verify the properties that f 0 (z) = f (z), f (0) = 1, f (z)f (w) = f (z + w), etc.
2. sin z, cos z 2
2
4
3
5
z z z z For f (z) = ez , look at f (iz) = 1+(iz)+ (iz) 2! +... = (1− 2! + 4! +...)+i(z− 3! + 5! +...). Then we set f (iy) =: cos y + i sin y. ∂ ∂ We can verify the usual properties that ∂z cos z = − sin z, ∂z sin z = cos z, cos 0 = 1, iz sin 0 = 0, cos(−z) = cos z, sin(−z) = − sin z, e = cos z + i sin z, etc.
Likewise, we can express cos z =
eiz + e−iz , 2 15
sin z =
eiz − e−iz . 2i
We can likewise check the addition formulas for cos(z1 + z2 ) and sin(z1 + z2 ). 3. log z w = log z ⇐⇒ z = ew . For w = a + bi, ew = ea (cos b + i sin b). Apparently, log z = log r+iθ+i(2πn), so the complex log is multi-valued. Write z = reiθ = r(cos θ+i sin θ). We define the principal branch of log z as follows. Consider C − R≤0 . log z has imaginary part in (−π, π). Write log z = log |z| + iArg(z) for Arg(z) ∈ (−π, π); this is our definition of log z for the principal branch, which has range −π < Im(z) < π. Claim. log z is holomorphic given a branch cut, and
∂ ∂z (log z)
= 1/z.
Proof. Consider lim
h→0
log(z + h) − log z . h
Set w1 = log(z + h), w = log z. Then, noting that log is continuous given a branch cut, we see that lim
h→0
w1 − w 1 1 1 log(z + h) − log z = lim w1 = ew1 −ew = w = . h→0 e h − ew e z limw1 →w w1 −w
Intuitively, log z looks like a helix above C − {0}; we call this the Riemann surface for log z. 4. z α , α ∈ R Define z α = eα log z = eα(log z+2πni) , which is potentially multi-valued. eα log z is the principal branch; now examine e2πnαi . If α ∈ / Q, e2πnαi takes infinitely many values. k 2πnki/l If α = l in reduced form, e takes l values (e.g. enπi = ±1 for α = 21 ). √ √ √ reiθ = reiθ/2 , For example, when considering z, we still need a branch cut. θ ∈ (−π, π). Integration. Given f (z) holomorphic, can we find another function F (z) holomorphic with 0 RF (z) = f (z)? The answer is, generally, yes. We begin with integration along a contour, f (z)dz for C a contour. C Definition. The image of γ(t) : [a, b] → C is a contour curve if: 1. γ is continuously differentiable except at finitely many points 2. When differentiable, γ 0 (t) 6= 0 except at finitely many points R Rb For C a contour, C f (z)dz = a f (γ(t))γ 0 (t)dt. Call γ1 (t) ∼ γ2 (t) for γ1 : [a, b] → C and γ2 : [c, d] → C if ∃λ(t) : [c, d] → [a, b] s.t. γ2 (t) = γ1 (λ(t)). Claim. This is well defined: Z b
f (γ1 (t))γ10 (t)dt =
Z
a
c
16
d
f (γ2 (t))γ20 (t)dt.
Lemma. (complex chain rule) Let g(t) = f (γ(t)), f (z) holomorphic. γ 0 (t)f 0 (γ(t)). Proof. As usual, with limits or whatnot.
Then g 0 (t) =
Example. Compute Z
z k dz
S1
for k ∈ Z, and S 1 the unit circle in C oriented counterclockwise. Explanation. Let γ(t) = eit = cos t + i sin t. By chain rule, γ 0 (t) = ieit . Then Z
Z
k
2π
e
z dz = S1
ikt
it
Z
ie dt = i
2π
ei(k+1)t dt.
0
0
If k = −1, this is Z S1
1 dz = i z
Z
2π
1dt = 2πi. 0
If k 6= 1, this is 2π
Z
i(k+1)t
e 0
2π ei(k+1)t = 0. dt = i(k + 1) 0
So z k for k ∈ Z, k 6= −1 has an antiderivative in C − {0}: log z, but it’s multivalued.
z k+1 k+1 .
1/z does not; we’d want
Proposition. If f (z) holomorphic with F (z) holomorphic and F 0 (z) = f (z), then Z f (z)dz = F (end) − F (start), C
for C a path from start to end. Proof. Use the complex chain rule.
Next time, we’ll consider functions that are defined on all of C (which we call entire functions), we’ll show ∃F (z) : F 0 (z) = f (z). Along the way, we’ll prove the closed curve theorem. For motivation, we recall Green’s theorem: Z b Z Z ∂Q ∂P 0 − dxdy (P, Q) · r (t)dt = ∂x ∂y a D for r(t) : [a, b] → R2 a closed curve around a region D. This requires that Qx , Py be continuous (or something like that). The theorem we’ll prove does not need this requirement.
17
CHAPTER
5
THE CLOSED CURVE THEOREM AND CAUCHY’S INTEGRAL FORMULA
Today we’ll be talking about the closed curve theorem and Cauchy’s integral theorem. Recall from last lecture that, for a parameterization γ of the contour C, we had Z Z f (z)dz = f (γ(t))γ 0 (t)dt. C
We also know that the path integral is determined by the antiderivative’s value at the curve endpoints: Proposition. If F (z) is holomorphic, F 0 (z) = f (z), then Z f (z)dz = F (final) − F (initial). C
The closed curve theorem. We will first provide motivation for the closed curve theorem. Given f and γ, consider splitting into real and imaginary parts: f (z) = f (x, y) = u(x, y) + iv(x, y)
Z
f (γ(t))γ 0 (t)dt = Z =
Z
γ(t) = a(t) + ib(t) Z ˙ ˙ + i(ub˙ + v a)]dt (u + iv)(a˙ + ib)dt = [(ua˙ − v b) ˙
˙ (u, −v) · (a, ˙ b)dt +i
Z
˙ (v, u) · (a, ˙ b)df
(5.1)
If C is the closed, counterclockwise boundary of D, then by Green’s theorem (which only applies when all partials are continuous) we have that the above is equal to Z Z Z Z ∂v ∂u ∂u ∂v − − dxdy + i − dxdy = 0, ∂x ∂y ∂x ∂y D D
18
where the last equality follows from the Cauchy-Riemann equations. We can also note that, in the language of vector fields, (1) is equivalent to Z Z Z Z curl(f )dA + div(f )dA, D
D
where f is thought of as a vector field on R2 induced by f . The equivalent formulation of the CR equations would then be curl(f ) = div(f ) = 0 as a vector field on R2 . The conclusion above gives us a glimpse into the closed curve theorem, which, however, does not require the first partials to be continuous. Keep in mind that our objective is the result that f being holomorphic on the interior of a disk implies that f has a convergent power series expansion on that disk: we will do this using the closed curve theorem and friends. Claim. The closed curve theorem (stated below) is true for linear functions f (z) = a + bz. R Proof. Let F (z) = az + 12 bz 2 , so F 0 (z) = f (z). So we have C f (z)dz = F (final) − F (initial) = 0. Theorem. (closed curve theorem) Suppose f (z) is holomorphic in an open disk D. Let C be any closed contour curve in D. Then Z f (z)dz = 0. C
Remark. Our strategy is in three steps: (1) show this for any polygonal curve; (2) show that f (z) has an antiderivative on D; (3) conclude the result. Proof for triangle. Let T be a triangle contour. The idea is that f being holomorphic implies that f is well-approximated by linear functions on small scales; see the claim above. R First, we subdivide the triangle. Suppose | T f (z)dz| = C. Then ∃Ti with boundary Γi R such that | Γi f (z)dz| ≥ c/4. Call this T (1) with boundary Γ(1) . Keep subdividing to get R (dropping the parentheses) T 1 , T 2 , T 3 , ... and Γ1 , Γ2 , Γ3 , .... such that | Γk f (z)dz| ≥ 4ck . k Note that ∩∞ k=1 T = {point}; call the point z0 . f (z) is holomorphic at z0 , so we have f (z) − f (z0 ) 0 lim − f (z0 ) = 0. z→z0 z − z0 Thus f (z) = f (z0 ) + f 0 (z0 )(z − z0 ) + (z)(z − z0 ) with (z) → 0 as z → z0 . Now consider Z Z Z f (z)dz = [f (z0 ) + f 0 (z0 )(z − z0 ) + (z)(z − z0 )]dz = Γk
Γk
(z)(z − z0 )dz.
Γk
Given > 0, ∃N s.t. for k ≥ N , (z) < for z ∈ T k . Let the perimeter of T be L and the perimeter of T k be 2Lk . The max distance between two points in T k would then be ≤ 2Lk . Thus Z 2 c ≤ L L = L , ≤ (z)(z − z )dz 0 k k k k 4 2 2 4 Γk
19
where the first term after the equality is length of Γk , theR second is an upper bound on (z), and the third is an upper bound on |z−z0 |. We had c = | Γ f (z)dz|, so c ≤ L2 =⇒ c = 0. Corollary. (extension to polygons) If P isRa polygon and f (z) is holomorphic on an open neighborhood of P with boundary Γ, then P f (z)dz = 0. Proof. Divide P into triangles. Theorem. (antiderivative theorem) If f (z) is holomorphic on an open disk D (or an open polygonally simply connected region Ω ⊂ C), then ∃F (z) holomorphic on D (or Ω) s.t. F 0 (z) = f (z). Definition. An open region Ω ⊆ C is polygonally connected (in topology, connected) if ∀z, w ∈ Ω, ∃ piecewise linear curve in Ω connecting z, w. A region is simply polygonally connected (simply connected) if any closed polygonal curve is the boundary of a union of polygons in Ω. Rz Proof of antiderivative theorem. Let p ∈ D (or Ω ⊆ C). Let F (z) = p f (z)dz (the line integral along any polygonal path from p to z, which is well-defined by the result on triangles above and the p.s.c. property of Ω). We check that Z z 1 F (z) − F (z0 ) = f (z)dz. z − z0 z − z0 z0 Note
because
Z z F (z) − F (z0 ) 1 − f (z0 ) = (f (z) − f (z0 ))dz z − z0 z − z0 z0 Z
z
Z
z
1dz = f (z0 )(z − z0 ).
f (z0 )dz = f (z0 ) z0
Then we have
z0
Z z 1 1 (f (z) − f (z0 ))dz ≤ |z − z0 |, z − z0 |z − z0 | z0
where the middle term after the inequality is the length of our curve and is upper bound for |f (z) − f (z0 )| (by continuity). So the LHS → 0 as → 0. Corollary. (closed curve theorem for regions) If f (z) is holomorphic on an open disk R D (or p.c.r. region Ω), then C f (z)dz = 0 for C being any R closed contour in D (or Ω). Proof. Take F (z), the antiderivative of f (z), so that C f (z)dz = F (final) − F (initial) = 0. Cauchy’s integral theorem. Given that f (z) is holomorphic, let’s look at ( f (z)−f (w) z 6= w z−w g(z) = 0 f (w) z=w for w ∈ C. Claim. g(z) satisfies the closed curve theorem. 20
Proof. g(z) is continuous because f 0 (w) exists. The goal is to prove the above result for triangles for g(z); then it will follow ∃G(z) : G0 (z) = g(z), along with the closed curve theorem for g(Z). Consider w. If w is outside T , then g(z) holomorphic in an open neighborhood of T and the old closed curve theorem applies. R If w ∈ Γ = ∂T , then we cut off small pieces {Ti } so Γi f (z)dz = 0 (because they do not contain w). The length of Γi can be arbitrarily small at M , the upper bound on g(z) on T , so it goes to 0 in the limit. If w is in the interior, we also use the same trick. Cauchy integral formula. This is one of the central results of complex analysis. Let C be a circle surrounding w ∈ C, and f (z) holomorphic on the closed boundary C of the open disk D. Then we have the formula 1 f (w) = 2πi
Z
f (z) dz z−w
Proof. Suppose C is a circle centered at w. The circle is parameterized by γ(t) = w+Reit for 0 ≤ t ≤ 2π. We know from the closed curve theorem that Z f (z) − f (w) 0= dz, z−w C because the integrand is g(z) defined on an interior satisfying the closed curve theorem. Now split this up: Z Z Z Z 2π f (z) 1 1 iReit dt dz = f (w) dz, dz = = 2πi Reit C z−w C z−w C z−w 0 Z 1 f (z) dz. =⇒ f (w) = 2πi z−w Theorem. Suppose f (z) is holomorphic on thePopen disk Dk (0) of radius R around 0 ∈ C. ∞ k 1/k Then there a convergent power series ≤ R1 such that k=0 ck z with lim|ck | P∞ exists k f (z) = k=0 ck z . Proof. From Cauchy’s integral formula, we have Z 1 f (w) f (z) = dw. 2πi w−z ˜ < R, and write Let |z| < R 1 1 1 = w−z w1− | wz | =
|z| ˜ R
z w
1 w
=
1 w−z
< 1 so this converges uniformly to 1 f (z) = 2πi
Z f (w)( ˜ C
1+
z z2 + 2 + ... w w
˜ Then for w ∈ C.
z z2 1 + 2 + 3 + ...)dz w w w
21
=
1 2πi
Z C
f (w) dw + w
i.e.
1 2πi
Z C
Z 1 f (w) f (w) dw z + dw z 2 + ..., w2 2πi C w3
1 Ck = 2πi
Z ˜ C
f (w) dw. wk + 1
We check this as follows: |Ck | ≤
1 1 ˜ = M/Rk , 2π RM ˜ k+1 2π R
˜ is the length of R and where the terms of the right of the inequality come as follows: 2π R M 1/k 1 1/k M is an upper bound on f in DR˜ (0). So lim|Ck | ≤ lim R˜ = R˜ , and f (k) (0) 1 = k! 2πi for C containing 0.
Z C
f (w) dw wk+1
Thus we have the result that f holomorphic =⇒ f analytic =⇒ f infinitely differentiable.
22
CHAPTER
6
APPLICATIONS OF CAUCHY’S INTEGRAL FORMULA, LIOUVILLE THEOREM, MEAN VALUE THEOREM
Applications of Cauchy’s integral formula. Recall that we have the Cauchy integral formula: Z 1 f (w) f (z) = dw, 2πi C w − z P for C a curve containing z. For f (w) = ck wk , note that we can write Z Z X Z ∞ 1 f (w) 1 c0 1 ck wk−1 dw = dw = dw = c0 . f (0) = 2πi C w 2πi C 2πi C w k=1
Let’s use this formula to compute some stuff. The general method is to decompose a closed contour (over which the integral is zero) into a sum of directed edges, then evaluating the integral over the other edges to give us results. Claim.
∞
Z 0
π sin x dx = . x 2
Proof. Let CR denote the contour of an upper half-circle of radius R centered at the origin. For r < R, we can use Cauchy’s integral theorem to write (noting that the integrand ix is equivalent to the imaginary part of ex : Z 0= r
R
eix dx + x iθ
Now parameterize Cr with z = re
Z CR
eiz dz + z
Z
−r
−R
eix − x
Z Cr
eiz dz. z
for θ ∈ [0, π] and likewise with z = Reiθ to see
Z π iReiθ Z −r ix Z π ireiθ eix e e e iθ 0= dx + iRe dθ + dx − ireiθ dθ. iθ iθ x Re x re r 0 −R 0 Z R ix Z π Z −r ix Z π iθ iθ e e = dx + ieiRe dθ + dx − ieire dθ. x r 0 −R x 0 Z
R
23
But note that the second integral → 0 as R → ∞ and the fourth integral → πi as r → 0 (show this yourself), so we’re done as sinx x (the imaginary part of the remaining two integrals) is symmetric over the y-axis and its integral from 0 to ∞ R ∞would then be equal to the imaginary part of our answer divided by 2. Namely, we have 0 sinx x dx = π2 as desired. Claim (added by Felix, for more practice) ∞
Z
sin(x2 )dx =
0
∞
−x2
e
2
e−x dx =
∞
∞
Z
e−(y
= −∞
−∞
√
2
−∞
Z
dx
2π . 4
cos(x2 )dx =
0
R∞
Proof. We first show that Z
√
∞
Z
2
π as follows:
+z 2 )
Z
2π
∞
Z
−∞
2
re−r drdθ = π,
dydz = 0
0
with y = r cos θ and z = r sin θ. 2 Using what we showed above, we integrate f (z) = e−z on the contour defined by C = {Reit : 0 < t < π/4}. Observe that we can bound the integral by π/4
Z Z 0 ≤ f (z)dz ≤
e−R
2
(cos2 (t)−sin2 (t))
0
C
π/4
Z
e−R
Rdt =
2
cos(2t)
Rdt,
0
R π/4 2 but also (from brute force bounding) we know that R 0 e−R cos(2t) dtR→ 0 as R → ∞ (∗). This is good, because we can invoke Cauchy’s theorem to see that C 0 f (z) for C 0 = {reit : 0 < r < R, 0 < t < π/4} = γ1 + γ2 + γ3 (where γ1 is the path from 0 to R, γ2 is the iπ/4 eighths-circle, and γ3 is the line from back to 0), a closed boundary, is zero. Letting R Re R → ∞, we saw from above that γ2 f (z) = 0, so what we have left is (I’ll drop the f (z)’s to make writing easier) Z Z Z ∞ Z Z Z √ √ 2 + = e−z + = π/2 + = 0 ⇒ π/2 = − . γ1
γ3
0
So to finish, we evaluate − Z −
Z =
γ3
R
e
γ3
R γ3
γ3
:
−(eiπ/4 t)2 iπ/4
e
γ3
dt = e
iπ/4
R
Z
e
0
−it2
=e
iπ/4
0
Thus we see that Z ∞ Z √ π/2 = eiπ/4 cos(t2 )−i sin(t2 )dt ⇒ 0
Z
R
cos(t2 ) − i sin(t2 )dt.
0
√
∞
cos(t2 )−i sin(t2 )dt = √
0
π/2 √ = 2/2 + 2/2i
R∞ R∞ Equating real and imaginary parts, we see that 0 sin(x2 )dx = 0 cos(x2 )dx = desired. (∗) Brute force bounding: For more exercise, we’ll explicitly show that Z R
π/4
e−R
2
cos(2t)
0
24
dt → 0
√
√ 2π 2π − i. 4 4
√
2π 4 ,
as
as R → ∞ here. We have Z π/4 Z 2 R e−R cos(2t) dt = R 0
π/4
e−R
2
sin(2t)
Z
π/6
e−R
dt = R
0
2
sin(2t)
Z
π/4
e−R
dt + R
0
2
sin(2t)
dt.
π/6
d Then, since we can bound sin x ≥ (1/2)x for 0 ≤ x ≤ 1/3 ( dx sin x ≥ 1/2 for x ≤ 1/3) and sin x ≤ 1/2 for π/6 ≤ π/4, we see that
Z
π/6
R
e−R
2
sin(2t)
Z
π/4
e−R
dt + R
2
sin(2t)
Z
2
Re−R t dt +
0
π/6
0
π/6
dt ≤
Z
π/4
Re−R
2
/2
dt
π/6
−πR2 2 π 1 Re−R /2 + (1 − e 6 ) → 0, 12 R as R → ∞. Integrals are fun stuff.
=
Now let’s turn to other theorems we can prove using the Cauchy integral formula. Theorem. (Liouville) A bounded entire function f (z) is constant. Proof. Suppose |f (z)| ≤ M for all z. Consider f (a), f (b) for |a|, |b| < R. We would like to show that f (a) = f (b). Let C0 (R) be the circle of radois R about 0. Then Z Z 1 f (z) f (z) |f (a) − f (b)| = dz − dz 2π C0 (R) z − a C0 (R) z − b and
f (z) (b − a)f (z) f (z) − = . z−a z−b (z − a)(z − b)
For |b − a| ≤ 2R, we can also get Z Z Z 1 f (z) f (z) 1 f (z)(b − a) 1 2πRM |b − a| dz − dz = dz ≤ 1 2 2π C0 (R) z − a z − b 2π (z − a)(z − b) 2π C0 (R) C0 (R) 4R =
4M |b − a| →0 R
as R → ∞. Let’s look at polynomials P (z) =
Pn
k=0 ck z
k
, cn 6= 0. What can we say about them?
Claim. limz→∞ P (z) = ∞ in the sense that ∀M, ∃r : ∀|z| > R, |P (z)| > M . Proof. Should be obvious.
Theorem. (Fundamental Theorem of Algebra) Given a nonconstant polynomial P (z) = Pn k c z , c n 6= 0, n ≥ 1, ∃w ∈ C : p(w) = 0. k=0 k 1 Proof. Suppose not. Then P (z) =: f (z) is a bounded entire function. Choose R such that n−1 X |ck | 1 > R |cn | k=0
25
and R > 1 so that for |z| > R we have 1 ≥ P (z)
1 1 2 R|cn |
Note that, in obtaining this inequality, we noted that the highest power of any polynomial dominates in the limit |z| → ∞. So f (z) is bounded by C for |z| > R, and it’s similarly bounded on any compact region D0 (R); thus f (z) is constant by Liouville’s theorem. Theorem. (Generalization of Liouville’s Theorem) If |f (z)| ≤ A + B|z|N and f (z) is entire, then f (z) is a polynomial of degree ≤ N . Proof. By induction. The base case is Liouville’s theorem. Now define ( f (z)−f (α) z 6= α z−α g(z) = 0 f (α) z=α for any α. Notice that g(z) is holomorphic, since we saw ∃G(z) that is holomorphic with G0 (z) = g(z), and we showed that holomorphic implies C ∞ so g(z) is also holomorphic. Notice further that |g(z)| ≤ C + D|z|N −1 . Addendum by Felix: We can see the last statement more clearly as follows. If to the contrary |g(z)| > D|z|n−1 , then taking limits as |z| → ∞ we see that the growth rate (α) or f 0 (α) (growth order N − 1). (growth order ≥ N ) would not match with f (z)−f z−a (Also, as said in lecture, we can see the last statement as follows: look inside a large ball, then outside the large ball. Inside, it’s bounded by some constant; outside, because of the induction hypothesis, it has the form D|z|N −1 .) So g(z) is a polynomial of degree ≤ N − 1 by induction and f (z) = g(z)(z − α) is a polynomial of degree ≤ N . This completes the inductive step. Uniqueness theorem. It turns out that holomorphic functions are determined up to their values in a certain region; this is the content of the uniqueness theorem. Definition. A region Ω in C is a polygonally connected open set. Proposition. (uniqueness theorem) If f and g are holomorphic on a region D ⊆ C with f (z) = g(z) on a collection of points accumulating somewhere, then f (z) = g(z) on D. Proof. Let A = {z ∈ D : ∃ infinitely many points w in any disk containing z s.t. f (w) = g(w)}. Let B = D\A. We claim that both A and B are open. To show that B is open, note that if z ∈ B, then ∃ a disk containing z with only finitely many such w’s. Let δ < limsuch w,w6=z |z|. Then every point in the open disk of radius δ is in B. To show that A is open, suppose z ∈ A. In D, ∃ a disk of radius r containing z. On that disk, both f and g being holomorphic implies that both f and g have convergent power series expansions centered at z of radius of convergence ≥ r. So by uniqueness for power series, f = g on this disk, i.e. every point in this disk in in A. So we have that A and B are both open, A ∪ B = D, and A ∩ B = ∅. By elementary topology, one must be the empty set. It turns out that polynomials are the only entire functions that go to infinity with z in ˆ with the limit. Geometrically, this means that if we consider the Riemannian sphere C 26
ˆ →C ˆ entire, then f : ∞ 7→ a for a finite implies that f (z) is bounded, i.e. a constant. f :C If f : ∞ 7→ ∞, then f (z) is a polynomial. Proposition. If f (z) is entire and f (z) → ∞ as z → ∞, then f (z) is a polynomial. Proof. The zeroes of f all lie in a disk of some radius, call it R. There are only finitely many zeroes, else they accumulate somewhere and then f = 0. Now consider g(z) =
f (z) . (z − z1 )m1 (z − z2 )m2 ...(z − zk )mk
P Note that g(z) has no zeroes and is entire. lim f (z) = ∞ =⇒ |g(z)| ≥ |z|1m , where m = mi . 1 So g(z) is entire and bounded, and by Liouville’s theorem it must be a constant. Thus f (z) is a polynomial. We take the time to state one last important result, the mean value theorem, without proof: Theorem. (mean value) If f (z) is holomorphic on Dα (R), then ∀r : 0 < r < R, we have Z 2π 1 f (α + reiθ )dθ. f (α) = 2π 0 Corollary. This is true as well for u = Re(f ), v = Im(f ) in place of f . Proof. Use the integral formula.
27
CHAPTER
7
MEAN VALUE THEOREM AND MAXIMUM MODULUS PRINCIPLE
Today, we’ll be talking about the maximum modulus principle, the mean value theorem, and some topological concerns. Recall from last time that we had the mean value theorem: Theorem. (mean value) For f holomorphic in DR (z), and r < R, Z 2π 1 f (z) = f (z + reiθ )dθ. 2π 0 We also have the maximum modulus principle, which is one of the most useful theorems in complex analysis: Theorem. (maximum modulus) Let f be holomorphic in DR (z). If |f | takes a local maximum value at z in the interior, then f is constant. Proof. Use the mean value theorem. Alternatively, an equivalent formulation (Ahlfors) is as follows: if f is holomorphic and nonconstant in DR (z), then its absolute value |f | has no maximum in DR (z). To show this, suppose that w = f (z) is any value in DR (z), so that the neighborhood D (w) is contained in the image of DR (z). In this neighborhood, there are points of modulus > |w| so |f (z)| is not the maximum of |f |. Claim. If |f | is constant and holomorphic, then f is constant on a disk (or region). Proof. Use the CR equations.
Corollary. (minimum modulus) If f is holomorphic and achieves a minimum of |f (z)| in a disk, then f (z) = 0 there. Proof. If f (z) 6= 0 at a minumum then 1/f is holomorphic in a neighborhood of z. Then use the maximum modulus theorem. Likewise, we can also use power series to prove this. We went over another proof of the maximum modulus theorem in class using power series; it may be nice to go over this proof as well (which is, again, rather long); see page 87 28
in the textbook. Corollary. If f is holomorphic in DR (z), then on Dr (z), r < R, the maximum value of |f (z)| on Dr (z) is achieved on the boundary. Proof. Consult the textbook. The following corollary is more general than the above: Corollary. If f is continuous on a closed bounded set E and holomorphic in its interior E − ∂E, then the maximum of |f | is attained on ∂E. Proof. (Ahlfors) Since E is compact, |f (z)| has a maximum on E. Suppose that it is attained at z0 ; if z0 ∈ ∂E, we’re done. Else if z0 is an interior point, then |f (z0 )| is also the maximum of |f (z)| in a disk |z − z0 | < δ contained in E. But this is not possible unless f (z) is constant in the component of the interior of E which contains z0 . It follows by continuity that |f (z)| is equal to its maximum on the whole boundary of that component. This boundary is not empty and is contained in the boundary of E, so the maximum is always attained at a boundary point. Proposition (“anti-calculus”) If |f (z)| achieves its max value on Dr (z) at w, then f 0 (w) 6= 0. Proof. Sketched in class; see page 88 of the textbook. We will use the maximum modulus principle to prove the following theorem: Theorem. (open mapping) The map of an open set under a non-constant holomorphic map f : U → C, U ⊆ C, is open. Note that f continuous at z means that ∀ > 0, ∃δ > 0 : f (Bδ (z)) ⊆ B (f (z)). Sketch of another proof. If f 0 (z) 6= 0, then ux uy Jf = v x vy is nonsingular. Then det Jf = ux vy − uy vx = u2x + vx2 = |f 0 (z)|2 6= 0. The inverse function theorem gives the result. If f 0 (z) = 0, then consider f (Bδ (0)) = Bf (δ) (0). Write f (z) = z k (ck + ck+1 z + ...). When f has a zero of order k at 0, i.e. f (0) = f 0 (0) = ... = f (k−1) (0) = 0, it turns out that f (z) = [g(z)]k , g(z) = Az + ... i.e. g 0 (0) 6= 0 (but we cannot prove this as of yet). Our proof. See page 93 of the textbook. Topology toward general closed curve theorem. We would like to take the time to revisit some topological concerns about the complex plane. In particular, recall that we defined the notions of polygonally connected and polygonally simply connected : Definition. A set U ⊆ C is polygonally connected if ∀p, q ∈ U , there exists any piecewise linear path from p to q. A polygonally connected set is connected, but a connected set may not be polygonally connected (what is an example?). For open sets, they are equivalent:
29
Proposition. If U ⊆ C is open, then U is connected iff it’s polygonally connected. Definition. A set U ⊆ C is polygonally simply connected (or, in the topological sense, simply connected ) if for any closed polygonal curve is the boundary of a union of polygons in U . Definition. A region Ω in C is an open connected (or polygonally connected) subset of C. If E(U ) denotes the space of directed edges (for purpose of integrating over) in the region U , then by composing elements in E(U ) to form a closed contour γ we know from R Cauchy’s integral theorem that γ f (z)dz = 0 for f holomorphic. The notion of connectedness goes into topology, involving such topics as homotopy and homology. For those of you that are familiar with topology, a simply connected set has trivial fundamantal group (e.g. every loop is homotopic to a constant). The Cauchy integral formula also has a more general, homology version involving winding numbers.
30
CHAPTER
8
GENERALIZED CLOSED CURVE THEOREM AND MORERA’S THEOREM
Today we’ll talk about the general closed curve theorem and Morera’s theorem. Theorem. (general closed curve theorem) If D is a region (open, polygonally connected ˆ − D is connected, then the closed curve theorem holds, i.e. subset) in C and C Z f (z)dz = 0 C
for f holomorphic on D and C a contour curve in D. ˆ ∼ ˆ with A ∪ B ⊃ S and Definition. A set S in C = S 2 is connected if @A, B open in C A ∩ B 6= ∅. The method for proving the general closed curve theorem is: ˆ − D is connected implies D p.s.c. 1. C 2. D p.s.c. implies closed curve theorem holds (see textbook for pictures). To show this, we’ll need to go back to last time: first let U be a region in C. Then define Pn • T (U ) := { i=1 ai Ti }, for Ti a triangle in U , ai ∈ Z, and any n. Pn • E(U ) := { i=1 ai Ei }, for Ei an edge in U (recall that an edge means a pair of points {P, Q} with P 6= Q). Pn • P (U ) := { i=1 ai Pi }, for Pi a point in U . • ∂T := E1 + E2 + E3 for T a triangle. • ∂E = Q − P ; ∂ : E(U ) → P (U ) and ∂ : T (U ) → E(U ). • e ∈ E(U ) is closed if ∂e = 0, e.g. e = E1 + ... + En is closed iff ±Ei form loops. Definition. e1 is homologous to e2 , denoted e1 ∼ e2 , if ∃t : ∂t = e1 − e2 . This is an equivalence relation. 31
Definition. U is p.s.c. if e is closed, e.g. e is null-homologous, which means e ∼ 0. Proposition. If U is convex in C, then U is p.s.c. Proof. Prove this yourself, by dividing the set into triangles.
Proposition. If U is p.s.c., then the closed curve theorem holds. Proof. Note that the antiderivative theorem holds. Let e be a (not necessarily closed) piecewise linear path from P to Q, ∂e = Q − P . Then Z X Z f (z)dz = ai f (z)dz. e
Define F (z) =
Rz z0
Ei
f (z)dz. This makes sense because Z Z f (z)dz = f (z)dz. e1
if e1 ∼ e2 , since then
e2
Z f (z)dz =
X
e1 −e2
Z bj
f (z)dz Γ
where Γ is the boundary of the triangle and the integral is zero. So the antiderivative theorem implies the closed curve theorem as before: Z f (z) = f (final) − f (initial) = 0, C
as desired.
We’ll introduce Morera’s theorem since we have some time. First some topology: Proposition. If U is an open subset in C, U is polygonally connected iff U is connected. ˆ If C ˆ − D is connected, then D is p.s.c. Claim. Let D be a region in C ⊂ C. Theorem. (Morera) If f (z) is continuous on a region D and then f (z) is holomorphic. Proof. See textbook.
R Γ
f (z)dz = 0 for Γ a triangle,
Morera’s theorem is quite useful. Some of the uses are as follows: 1. Showing limn→∞ fn = f is holomorphic. 2. Showing that some functions defined by sums or integrals are holomorphic, e.g. the Riemann zeta function and the gamma function: ζ(z) := Z Γ(z) := 0
32
∞ X 1 z n n=1
∞
e−t tz−1 dt
3. Showing functions that are continuous and holomorphic on all but some set (e.g. some point) are holomorphic. Proposition. Suppose {fn } is holomorphic in an open set D, and fn → f uniformly on compact subsets of D. Then f = limn→∞ fn is holomorphic. Proof. It suffices to work in B (α) ⊂ D. B/2 (α) is compact, so by assumption fn → f is uniform here. Note that fn → f uniformly implies f is continuous. Also Z Z Z f (z)dz = lim fn (z)dz = lim fn (z)dz = 0 Γ n→∞
Γ
n→∞
Γ
R
(Uniform convergence implies that we can swap and lim, as we all know from elementary analysis.) By Morera’s theorem, f is holomorphic. The Riemann zeta function. The zeta function ζ(z) is defined as ζ(z) =
∞ X 1 , nz n=1
for Re(z) > 1. We claim that ζ(z) is convergent for Re(z) > 1, and indeed, is uniformly convergent for Re(z) > R > 1. (Hence by Morera’s theorem ζ(z) is holomorphic.) Proof of claim. First note that n1z = e−(log n)z . We first show uniform convergence. Let fN (z) := We bound
P∞
n=N +1
N N X X 1 = e−(log n)z . z n n=1 n=1
|e−(log n)z | for the whole region Re(z) ≥ R > 1: |e−(log n)z | = |e− log n eRe(z) | ≤ R|e− log n |
P∞ So we’re left with n=N +1 the tail end is small.
1 , nR
R > 1. The integral test enures convergence, telling us that
33
CHAPTER
9
MORERA’S THEOREM, SINGULARITIES, AND LAURENT EXPANSIONS
Let’s finish the topology stuff, talk about Morera’s theorem, and then move on to singularies and Laurent expansions. ˆ = S2? First, what does it mean to fix an open set in C ˆ is Definition. A neighborhood of ∞ is given by {∞} ∪ {z : |z| > R}. Then S ⊂ C open if: (1) S ∩ C is open in C (2) If ∞ ∈ S, then ∃R : {z : |z| > R} ⊂ S. Lemma. If P is a closed polygonal path which doesn’t cross itself (i.e. P is a piecewise linear map from ˆ s.t. A∪B = C−P ˆ an interval that is injective except at the endpoints), then ∃A, B open in C and A ∩ B = ∅. Proof. Note ∃R : P ∩ {|z| > R} = ∅ for z ∈ C − P . Take L, a line segment from z to a point in S = {|z| > R}, and count the number of times P crosses L to R and the number of times P crosses R to L. This makes sense because P ∩ L is a finite union of edges of P and points. Let z ∈ A if the count is odd; else z ∈ B. Exercise. Show that z ∈ A or B is well-defined independent of the choice of L. Hint: use wedges. Conclude that nearby points are in the same set. Integration over other regions. We use C − R≤0 to denote the complex plane with ˆ − (C − R≤0 ) = {∞} ∪ R≤0 is connected. a ray on the real axis missing. Note that C Proof: any interval is connected. Thus we see that C − R≤0 is polygonally simply connected (PSC), and we can integrate along any closed curve in this region by applying the closed curve theorem. How about C∗ = C − {0} ? Applying the closed curve theorem for the difference contour
34
C2 − C1 , we see that
Z f (z)dz = 0 C2 −C1
for f (z) holomorphic on C∗ . An image of C2 (the outside counterclockwise contour) and C1 (the inside counterclockwise contour) is as follows:
Diagram 1: C1 and C2 in C∗ . Now let’s pick off with Morera’s theorem, which as you recall states Theorem. (Morera) If f is continuous on an open set D and ing the boundary of a triangle in D, then f (z) is holomorphic.
R Γ
f (z)dz = 0 for Γ be-
Corollary. If fn (z) is holomorphic on D open and fn → f uniformly on compact subsets of D, then f (z) is holomorphic on D. Proof. Proved last time. The Riemann zeta function. The Riemann zeta function is a particular form of a Dirichlet series or L-function given by ζ(z) =
∞ X 1 . nz n=1
We claim that ζ(z) is holomorphic on Re(z) > 1, and indeed we showed this last time. The gamma function. The gamma function, which is useful in probability and number theory, is given by Z ∞
e−t tz−1 dt.
Γ(z) = 0
We claim that Γ(z) is holomorphic for Re(z) > 0. Remark. Γ(z) is a natural extension of the usual factorial function in that Γ(n + 1) = n!. Show this yourself using integration byRparts and induction. ∞ Proof of the claim. We show that 0 e−t tz dt is holomorphic in Re(z) > −1. To that extent we work in −1 < R1 ≤ Re(z) ≤ R2 ; note that any compact subset of {z : Re(z) > −1} is contained in one of these sets. Define Z 1 Z n fn (z) := e−t tz dt + e−t tz dt, 1/n
1
which we will show converges uniformly. For t ≥ 1, we can bound |e−t tz | ≤ e−t tRe(z) ≤ e−t tR2 because tz = ez log t = e(Re(z)) log t eiIm(z) log z , 35
the last term of which has norm 1. Then Z
∞
|fn (z)/2 − f (z)/2| ≤
e−t tR2 dt,
n
which converges for n = 1. So ∀ > 0, ∃N : ∀n > N , Z ∞ e−t tR2 dt < /2. n
For t ≤ 1, |e−t tz | = e−t tRe(z) ≤ e−t tR1 and Z Z Z Z 1 1/n 1 1/n −t z −t z −t z e−t tR1 dt e t dt ≤ e t dt − e t dt = 0 1/n 0 0 R∞ for R1 > −1. Let s = 1/t and ds = − t12 dt, so that we get n e−1/s s−2−R1 ds. We need −2R− R1 < −1, i.e. −1 < R1 . Thus the sequence converge uniformly, so ∀ > 0, ∃N : ∀n > ∞ N, n e−1/s s−2−R1 ds < /2. Corollary. If f is continuous on D and f is holomorphic on D except at α ∈ D, then f is holomorphic on D. Proof. Let Γ = ∂T . Then Z f (z)dz = 0 Γ
if 0 ∈ / T . If α ∈ T , we split the contour so that α is located at the corners of the triangle. If α is at a corner, then Z Z 0 = lim f (z)dz = f (z)dz l→∞
Γl
Γ
by continuity of f .
Singularities of functions. (Newman and Bak §9) Singularities of functions will be important when trying to access their behaviors around certain points so that we can define things like Laurent series or perform residue calculus. Needless to say, an understanding of the different types of singularities and how they pertain to classes of functions is important for complex anaysis in general. Definition. Given α ∈ C, a deleted neighborhood D of α is a neighborhood −{α}, e.g. {z : 0 < |z − α| < }. f is said to have an isolated singularity at α if f is defined and holomorphic on a deleted neighborhood of α. Types of singularities. For D a deleted neighborhood of α, we have the following types of singularities: 1. Removable singularity: ∃g(z) on D ∪ {α} holomorphic with g(z) = f (z) on D. 2. Pole of order k: ∃A(z), B(z) holomorphic on D ∪ {α} s.t. A(α) 6= 0, B(α) = 0, A(z) f (z) = B(z) and B(z) has a zero of order k at α. Recall that we can expand B(z) into a power series as B(z) = ck (z − α)k + ck+1 (z − α)k+1 . 3. Essential singularity: Neither of the above.
36
The following proposition allows us to recognize if something has a removable singularity. Proposition. If lim f (z)(z − α)
z→α
exists and is equal to 0, then f (z) has a removable singularity at α. Proof. Define h(z) as ( f (z)(z − α) z 6= α h(z) = 0 z = α. This is continuous, and holomorphic except at α. So h(z) is holomorphic in a neighborhood h(z) of α by the corollary to Morera’s theorem. Moreover, we see that h(α) = 0, so g(z) = z−α is holomorphic and equal to f (z) away from α. (If h(α) = 0 and h is holomorphic, then h(z) z−α is holomorphic; one can prove this easily using power series.) Likewise, the following proposition allows us to recognize a pole of f . Proposition. Suppose f (z) is holomorphic in a deleted neighborhood of α and ∃n : lim f (z)(z − α)n = 0.
z→α
Then letting k = 1 be the least such n, f (z) has a pole of order k at α. Proof. Again, let ( f (z)(z − α)k+1 z 6= α h(z) = 0 z = α. h(z) is continuous on a neighborhood and holomorphic on a deleted neighborhood, so it’s holomorphic on a neighborhood of α again by the corollary to Morera’s theorem. h(α) = 0 h(z) g(z) so g(z) = z−α is holomorphic in D∪{α} and g(z) = f (z)(z −α)k on D. Then f (z) = (z−α) k. (Note limz→α g(z) 6= 0 by assumption; also notice it exists). For essential singularities, we’ve shown that it must be the case that limz→α f (z)(z − α)n does not exist for all n. Notice as well the following: • If f is bounded in a deleted neighborhood, then f has a removable singularity. • If f < C1 |z|1N + C2 |z|N1 −1 + ... + C as z → 0, then f has a pole of order ≤ N . We have a theorem for essential singularities, which tells us that the image of any deleted neighborhood of an essential singularity under a holomorphic function is necessarily dense in the complex plane: Theorem. (Casorati-Weierstrass) If f has an essential singularity at α and D is any deleted neighborhood of α, then f (D) = {f (z) : z ∈ D} is dense in C (i.e for any > 0, w ∈ C, B (w) ∩ f (D) 6= ∅). Proof. Suppose not. Then ∃B (w) with B (w) ∩ f (D) = ∅. This means |f (z) − w| > , or 1 1 1 |f (z)−w| < . So f (z)−w is defined on D, a deleted neighborhood of α, and bounded there. 1 Thus g(z) = f (z)−w is holomorphic on a neighborhood of α. 1 = f (z) − w, so that f (z) = wg(z)+1 is a ratio of two holomorphic Now consider g(z) g(z) functions; we see that the singularity must be removable, or f (z) must have a pole.
37
Exercise left to the reader: prove the converse of the theorem.
Example. Consider f (z) = e1/z defined on C ∗ , which has an essential singularity at 0 ∈ C. We claim that f (B (0) − {0}) is C∗ . To see this, note that f (z) is the composition of z1 and then ez . Under z1 , B (0) − {0} gets inverted to fill C outside the ball. ez is 2πi-periodic, so we can draw horizontal lines in the complex plane at π + 2πn, n ∈ Z. Then the interior of any strip created by these horizontal lines gets mapped under ez to C − R≥0 . Our claim is that there are strips outside B0 (1/) so that f (B (0) − {0}) = C∗ .
Diagram 2: The mappings z1 : B (0) − {0} → C − (B (0) − {0}) and ez : {z : π ≤ Im(z) ≤ 3π} → C − R≥0 . Laurent expansions. P∞ f (z) = k=−∞ ck z k .
Simply put, Laurent expansions are Taylor series of the form
Theorem. If f (z) is holomorphic on the annulus A(R1 , R2 ) := {z : R1 < |z| < R2 }, P∞ then f (z) = k=−∞ ck z k converges on all of A(R1 , R2 ). (Taking R1 = 0 and R2 = ∞ is also fine.) We say L.
P∞
k=−∞
ak = L if
P∞
k=0
ak exists,
P−1
−∞
ak =
P∞
l=1
a−l exists, and their sum is
Proposition. If 1 limk→∞ |ck |1/k P∞
≥ R2 and lim |c−k |1/k ≤ R1 , k→∞
k
then f (z) = k=−∞ ck z converges and is holomorphic on A(R1 , R2 ). Proof. Write k ∞ ∞ X X 1 f (z) = ck z k + c−k . z k=0
The first sum is convergent for |z| ≤ | z1
≤
1 limk→∞ |c−k |1/k
=⇒ |z| ≥
k=1
1
limk→∞ |ck |1/k limk→∞ |c−k |1/k . The
. The second sum is convergent for
first sum is clearly holomorphic. The P∞ second sum is holomorphic as a function of 1/z; e.g. g(z) = k=1 c−k wk is holomorphic 1 1 for |w| ≤ R1 . The second sum is a composition of z and g, so by the chain rule it’s also holomorphic. So f is holomorphic.
38
Example. The following Laurent series expansion converges on A(0, ∞): e1/z = 1 +
1 1 1 + 2 + 3 + ... z 2z 6z
Example. Near z = 0 (pole of order 1), we have 1 1 1 1 = = (1 − 2z + 3z 2 − 4z 3 + ...). z(1 + z)2 z (1 + z)2 z This is because near z = −1.
1 ∂ ∂z ( 1+z )
1 = − (1+z) 2 and
1 1+z
= 1 − z + z 2 − z 3 + .... We can do the same
39
CHAPTER
10 MEROMORPHIC FUNCTIONS AND RESIDUES
Last time, we introduced the Laurent series for f (z) as f (z) = defined on Aα (R1 , R2 ), the annular region centered at α.
P∞
Theorem. If f (z) is holomorphic on Aα (R1 , R2 ), then f (z) = the series converges on Aα (R1 , R2 ). Proof. Do this yourself by considering the function ( f (w)−f (z) w 6= z w−z g(w) = 0 f (w) w=z and using the closed curve theorem.
−∞ ck (z
P∞
− α)k . This is
k=−∞ ck (z
− α)k and
Remark.P Laurent series are P unique to every function. We define the principal part of −1 f (z) = near α as the sum k=−∞ ck (z − α)k and observe the following: (1) α is a removable singularity iff ck = 0 for k < 0. (2) α is a pole of order n iff ck = 0 for k < −n, c−n 6= 0. (3) α is an essential singularity iff ck 6= 0 for infinitely many negative k (why?). Here’s an idea: at a pole, we can make the function f (z) still be holomorphic if we exˆ so that f (z) = ∞ there and holomorphically so. pand its range to C, 1 Proposition. If f (z) has a pole of order k at α, then f (z) is holomorphic near α and has a pole of order k. 1 1 Proof. Write f (z) = (z−α) k g(z). Then g(z) holomorphic near α, g(z) 6= 0, so f (z) = (z−α)k g(z) ,
g(α) 6= 0. This is holomorphic on a neighborhood at α and has zero of order k.
Definition. f is meromorphic on D if f (z) is holomorphic on D except at isolated singularities at which f has poles. The proposition above can then be restated as: a function f being meromorphic in D ˆ is holomorphic. really means that f : D → C 40
Theorem. Let f be meromorphic in C and at infinity, and suppose that limz→∞ f (z) = ∞ for some w ∈ C. (The limit means ∀M, ∃R : |f (z)| > M when |z| > R.) Then f (z) is a rational function. ˆ →C ˆ is holomorphic, then f is a rational function. Theorem. If f : C Residues of functions. By introducing residues, we seek to deduce the residue theorem as a means for computing integrals more effciently. Given a holomorphic P∞function f (z) in a deleted neighborhood of α, by the Laurent expansion we have f (z) = −∞ ck (z − α)k . We define the residue of f (z) at the point α, Res(f ; α), as Res(f ; α) = c−1 . From Cauchy’s integral formula, we notice that Z 1 f (w)dw = c−1 . 2πi CR (α) Also recall the expression for ck : ck =
1 2πi
Z CR (0)
f (w) dw. wk+1
We see from this that the “limsup stuff” shows that Laurent series are uniformly convergent on A(r1 , r2 ) by comparison with geometric series. On a ball of radius , we can compute the residue of f at α by using the formula: Z Z 1 1 Res(f ; α) = c−1 = f (w − α)dw = f (w)dw. 2πi C (0) 2πi C (α) There are a few ways to compute residues in general: R (0) Compute the integral C (α) f (w)dw. (1) Compute the Laurent series for f centered at α. (2) If f has a simple pole, then c−1 = lim (z − α)f (z). z→α
Also, if f (z) = A(z)/B(z) for A a nonzero function, and B having a simple zero, then we have: Lemma. Res(f ; α) =
A(α) B 0 (α) .
We finish by working out an example. Consider Z ∞ 1 dx. 2 −∞ 1 + x We can compute this by first taking a semicircle contour of radius R, for which there is one pole contained inside (α = i). If we compute the residue at this pole, we can use the residue theorem to get the value of the integral (this is left to the reader as an exercise). 41
CHAPTER
11
WINDING NUMBERS AND CAUCHY’S INTEGRAL THEOREM
Today, we’ll cover residues, winding numbers, Cauchy’s residue theorem, and possibly the argument principle. This is in the textbook, §10.1, §10.2. Last time, we defined resudies. For f holomorphic in a deleted neighborhood of α, we have Z 1 Res(f ; α) = f (z)dz = c−1 , 2πi C (α) P∞ where c−1 appears in the Laurent expansion for f around α, f = −∞ cn z n . Note that f is holomorphic in C (α) except for the point at α. 1 Example. Res( 1+z 2 ; i) Method 1: Find the Laurent expansion about i, i.e. in terms of z − i.
1 −i 1 1 1 1 = = + − (z − i) + .... 1 + z2 (z + i)(z − i) 2 z−1 4 8 So c−1 = −i 2 . Method 2: If f (z) has a simple pole at α, i.e. if f (z) = 0, B 0 (α) 6= 0, then Res(f ; α) =
A(α) B 0 (α) .
Set
1 Res( 1+z 2 ; i)
=
1 2i
A(z) B(z)
for A(α) 6= 0, B(α) =
= −i/2.
An application of this is as follows. Consider the integral Z ∞ 1 dx = lim arctan(R) − lim arctan(−R) = π/2 − (−π/2) = π. 2 R→∞ R→∞ −∞ 1 + x We can also do this by contour integration over the contours C1 , C2 , C3 , where C1 = [−R, R], C2 is the CCW semicircle connecting R to −R, and C3 is the CCW circle omitting the point at i. Then Z f (z)dz = 0 C1 +C2 −C3
42
by the closed curve theorem, and by the estimation lemma we have Z 1 1 π dz ≤ πR · 2 = →0 2 R R C2 1 + z as R → ∞. Also, Z C3
1 1 dz = 2πiRes( ; i) = 2πi(−i/2) = π. 1 + z2 1 + z2
So Z
R
lim
R→∞
−R
1 dx = π 1 + x2
after substituting for the values of the integrals over C1 , C2 , C3 . Winding numbers. These intuitive give the number of times a path loops around a point, and are also dealt with in topology. Definition. Let γ : [0, 1] → C be a contour curve with γ(t) 6= α ∀t. Then the winding number of γ about α, denoted n(γ, α), is the value Z 1 1 dz. 2πi γ z − α Theorem. n(γ, α) ∈ Z. Proof. We have Z γ
1 dz = z−α
Z 0
1
γ 0 (t) dt. γ(t) − α
d dt
Note that the integrand on the RHS is log(γ(t)−α), with a caveat that log is a multivalued function. log z = log |z| + iArg(z) modulo 2πi. We’ll show that Z 0
1
γ 0 (t) dt = 2πik γ(t) − α
for some k. This k = n(γ, α). Let F (s) = 0
Then F (s) =
γ 0 (s) γ(s)−α .
γ 0 (t) dt 1 γ(t)−α
Rs
(secretly, log(γ(s)−α)−log(γ(1)−α).
We claim that F (s) =
γ(s) − α . γ(0) − α
To see this, let G(s) = (γ(s) − α)e−F (s) . Then G0 (s) = −γ 0 (s)e−F (s) + γ 0 (s)e−F (s) = 0, so G(s) is a constant. Check that at s = 0, we get G(0) = (γ(0) − α)e−F (0) , so indeed eF (0) = γ(0)−α γ(t)−α . R 1 γ 0 (t) F (1) Now eF (1) = γ(1)−α = 1. So F (1) = 2πik γ(t)−α . F (1) = 0 γ(t)−α dt and γ(1) = γ(0), so e for k ∈ Z. Corollary. (version of Jordan curve theorem) Let γ : [0, 1] → C be a closed curve. Then C − image(γ) is disconnected. 43
Let’s call a simple closed curve one that doesn’t intersect itself. For us, if γ has n(γ, α) = 0 or 1, then we call γ “winding simple.” The Jordan curve theorem shows that any closed curve has what we call an interior and exterior. For the interior region B, we would have n = 1. For the exterior region A, we would have n = 0. Cauchy’s residue theorem. It turns out that we can evaluate any contour integral of a holomorphic function by considering its residues and winding numbers. Theorem. Let f be holomorphic on a region D with zero first homology (i.e. PSC, i.e. closed curve theorem holds for holomorphic functions on D) except for isolated singularities at α1 , ..., αn ∈ D. Let γ be any contour curve in D. Then Z f (z)dz = 2πi γ
n X
n(γ, αk )Res(f ; αk ).
k=1
Proof. The gist is to repeatedly subtract the principal parts of f at each αk . This proof is a bit long to type up, so it is left as an exercise to the reader.
44
CHAPTER
12 THE ARGUMENT PRINCIPLE
Recall from last time that we had Cauchy’s residue theorem: ˆ − D connected (i.e. closed curve theorem applies), i.e. Theorem. Let D be a region with C simply connected (all loops contract to a point in D). Let f be holomorphic on D except at α1 , ..., αn , and let γ be a contour curve in D missing these points. Then Z f (z)dz = 2πi γ
n X
Res(f ; αk )n(γ, αk ).
k=1
The argument principle. This is essentially relating the number of zeroes and poles of f to (# of times f winds around 0). Let γ be a regular contour curve, e.g. if ∀α ∈ C − im(γ), either n(γ, α) = 0 or n(γ, α) = 1. Intuitively, this means that the points are either “inside of γ” or outside; the former is given by the set {z ∈ C − im(γ) : n(γ, z) = 1} and the latter is given by the set {z ∈ C − im(γ) : n(γ, z) = 0}. Theorem. If f is holomorphic in a simply connected region D, and γ a regular curve in D, then the following are equal: (1) (# of zeroes of f with multiplicity) − (# of poles of f with multiplicity inside γ) (2) The winding number around zero of f (γ(t)), i.e. n(f ◦ γ, 0) R f 0 (z) 1 (3) 2πi dz γ f (z) Remark. This assumes that for f |γ 6= 0, there are no poles. The multiplicity is also called the order. Proof. Let’s do (2) ⇐⇒ (3). For γ : [0, 1] → C, we have 1 n(f ◦ γ; 0) = 2πi
Z f ◦γ
1 1 dz = z 2πi
Z
1
0
γ 0 (t)f 0 (γ(t)) 1 dt = f (γ(t)) 2πi
Z γ
f 0 (z) dz. f (z)
∂ The integrand is actually ∂z (log f ), which keeps track of how Arg(f ) changes and goes around γ. (3) ⇐⇒ (1). To do this we check the residues of f 0 /f . The only singularities of f 0 /f occur at the zeroes or poles of f . Near a zero or pole α of f , we have f (z) = (z − α)n g(z)
45
for g(z) holomorphic near α, g(α) 6= 0. Also, f 0 (z) = n(z − α)n−1 g(z) + (z − α)n g 0 (z), so
f 0 (z) n g 0 (z) = + , f (z) z−α g(z)
n where we note that the residue at α of z−α is n, and g(z) 6= 0 implies that the term on the right is holomorphic and is the residue at α. Hence Z 0 X X 1 f dz = n(γ, α) · order(f, α) − n(γ, α) · order+poles(f, α) 2πi γ f α pole of f outside γ
α pole of f inside γ
And n(γ, α) → 1 in all these sums.
Moreover, we can say that the “zeroes of order n look like z n ”; they wrap around 0 n times CCW, and “the poles of order n look like z1n ”; they wrap around 0 n times CW. The idea of computing the number of zeroes in a curve by computing an integral is quite nice. Most of the time, however, we just use the following corollary (also dropping the idea of poles for now, since they aren’t really used in it): Corollary. (Rouch´e Theorem) (Assume f has no zeroes on γ.) Let f, g be holomorphic in the unit disk (in γ), and ||g(z)|| < ||f (z)|| on the unit circle (on γ). Then # of zeroes of f in the unit disk = # of zeroes of f + g in the unit disk, for γ regular. Examples. The # of zeroes of 3z 8 (+iz 6 + 1) in the unit disk is the same as 3z 8 , which has 8 zeroes. iz 8 + 3z 6 + 1 has 6 zeroes in the unit circle. iz 8 + z 6 + 3 has no zeroes. Proof of Rouch´e’s Theorem. First note that for any functions A and B, A0 B0 (AB)0 = + AB A B Now write
(f + g) = f
g 1+ f
,
and note | fg | < 1 on γ. The # of zeroes of f + g in γ is given by 1 2πi
Z γ
(f + g)0 1 dz = f +g 2πi
Z γ
f0 1 dz + f 2πi
Z (1 + g )0 f γ
(1 + fg )
dz,
but this is just g◦γ = (# of zeroes with multiplicity of f in γ) + n 1 + ;0 . f ◦γ Note that 1 + fg◦γ ◦γ is contained inside B1 (1) ⊆ {Re(z) > 0}, the upper half-plane. So the winding number is zero, and the above is Z 1 1 = dz 2πi ρ z 46
where ρ is the curve traced out by the fact that 1 + given by the principal branch of log z in Re(z) > 0.
g◦γ f ◦γ
< 1, which has an antiderivative
...................................................................................... Generalized Cauchy’s integral formula. Let γ be regular, and z be inside γ. Then Z k! f (w) (k) f (z) = dw. 2πi γ (w − z)k+1 Proof. The residue of
f (w) (w−z)k+1
at zero is f (k) (z) f (w) 1 = (w − z)k f (w) = . k+1 w − z (w − z) k!
Corollary. Suppose fn is holomorphic and fn → f uniformly on compact sets (hence holomorphic by Morera’s theorem). Then fn0 → f 0 uniformly as well. Proof. We have Z 1 fn (w) − f (w) fn0 (z) − f 0 (z) = dw 2πi γ (w − z)2 for γ = Cr (α), z ∈ Dr (α). On Dr/2 (α), we have Z 1 1 fn (w) − f (w) 4 0 0 |fk (z) − f (z)| = dw ≤ · 2πr · 2 · max |fn (w) − f (w)|, 2π γ (w − z)2 2π r Cr (α) where the r42 comes as an upper bound on Now choose n large enough so that
1 |w−z|2
because |w − z| ≥ r/2 =⇒
|fn (w) − f (w)|
N . Now cover the compact set in euqation by disks of half the radius. Because the set is compact, finitely many suffiice. Take the largest Nα among these. Theorem. (Hurwitz) Suppose fn → f holomorphic in D, and uniformly so on compact sets. Suppose that none of the fn ’s are zero in D. Then either f = 0 on D or f has no zeroes. Proof. Suppose f 6= 0, and let f (α) = 0 for some α ∈ D. α is not the limit of zeroes of f (other than 0), because then f = 0 by the uniqueness theorem. Hence there is some closed disk D (α), > 0, with no zeroes of f in D (α) other than α. We have fn0 → f 0 uniformly, and there are no zeroes on C (α) of f ; none of fn are zero, so f1n → f1 uniformly on C (α) (show this yourself). Thus fn0 f0 → fn f uniformly on C (α). It follows that Z Z 1 fn0 1 f0 lim = . n→∞ 2πi C fn 2πi C (α) f (α) 47
So lim # of zeroes of fn in D (α) = # of zeroes of f in D (α) = 1,
n→∞
a contradiction, so we can’t have this isolated zero of f .
48
CHAPTER
13 INTEGRALS AND SOME GEOMETRY
Recall the residue theorem: let αk be the singularities of f . Then Z X f (z)dz = 2πi Res(f ; αk )n(γ, αk ). γ
There are many applications of this. If P, Q are polynomials, Q(x) 6= 0 for real x, and deg Q ≥ deg P + 2, then Z
∞
−∞
n
X P (x)dx dx = 2πi Res Q(x)dx k=1
P (z) , αk . Q(z)
You can show this with the usual residue calculation (yes, the ones that we beat to death in our review session...). Example. Z
∞
−∞
1 dx x6 + 1
The poles are where z = −1, e.g. at e for θ = π6 , 3π 6 , .... Recall that, for f (z) = A(z)/B(z) with B(α) = 0, A(α) 6= 0, if B 0 (α) 6= 0 then this is a simple pole, and the residue is given by Res(f ; α) = A(α)/B 0 (α). So using this we get that the residues are 6
iθ
Res(f (z); eiπ/6 ) =
1 6e5πi/6
1 6eπi/2 1 Res(f (z); e5iπ/6 ) = πi/6 6e Res(f (z); eiπ/2 ) =
Then
Z
∞
−∞
x6
1 2πi −5πi/6 2π dx = (e + e−iπ/2 + e−iπ/6 ) = . +1 6 3 49
Geometry of complex functions. Recall the open mapping theorem: if f is holomorphic and nonconstant, and if U is open in C, then f (U ) is open, i.e. given , ∃δ : Bδ (z) ⊆ f −1 (B (w)) = {x ∈ C : f (x) ∈ B (w)}, i.e. |z − α| < δ =⇒ |f (z) − f (w)| < δ. Open means that ∀δ, ∃ : f (Bδ (z)) ⊇ B (w), so |f (z) − β| < =⇒ ∃α : |z − α| < δ, f (α) = β. Proof. Let α ∈ C. Wlog f (α) = 0; take C inside Bδ (α). Take C inside Bδ (α); min f (C) exists and it’s not zero, so call it z . Let w ∈ B (0). Then for z ∈ C, |f (z) − w| ≥ |f (z)| − |w| ≥ 2 − = . For z = α, |f (α) − w| = | − w| < . Hence min |f (z) − w| for z ∈ Bδ/2 (α) is not achieved on the boundary. Thus, by the minimum modulus theorem, ∃ a zero, e.g. f (z) − w = 0 in B. Theorem. (Schwarz Lemma) Let D be the unit disk. If f : D → D is holomorphic (extending continuously to the boundary) and f (0) = 0, then ( |f (z)| ≤ |z| |f 0 (0)| ≤ 1 with equality in either iff f (z) = eiθ z. Proof. Define ( g(z) =
f (z) z 0
f (0)
z= 6 0 z = 0,
(z)| which is holomorphic on the circle of radius δ; |g(z)| = |f|z| ≤ 1δ . By the maximum modulus principle, we know in fact |g(z)| ≤ 1 for all z ∈ D. Let δ → 1. So if a max is achieved on the interior, then f (z) is constant, i.e. g(z) is constant if |f (z)| = |z| for any z ∈ D1 (0), z 6= 0, or |f 0 (0)| = 1. This means g(z) = eiθ ; then g(z) = f (z)/z (f 0 (0)), so f (z) = eiθ z.
Corollary. If a map from D → D, f (0) = α, has a maximum value of |f 0 (0)| among maps D → D, then f is surjective. Proof. Consider h(z) = Bα (f (z)). h(0) = f 0 (0)Bα0 (0), h(z)D → D, h(z) : 0 7→ 0, |h0 (0)| ≤ 1. The conclusion is that 1 . |f 0 (1)| ≤ 0 Bα (α) = 1 − |α|2 Notice that this is achieved for inverse of Bα , which is Bα . Bα (Bα (z)) = z. Also note that Bα (z) =
z−α , 1 − αz
Bα0 (z) =
1 , 1 − |α|2
and
so Bα0 (0) = 1 − |α|2 , Bα (0) = 0, |Bα (z)| ≤ 1 for |z| ≤ 1. We used h0 (0) ≤ 1. We also know |h(z)| ≤ |z| and h(z) = Bα (f (z)); |Bα (f (z))| ≤ |z|. 50
We can therefore conclude that |f (z)| ≤ |Bα (z)| if f (0) = α, f : D → D. Riemann mapping theorem. (casually stated) Given a simply connected region U ⊆ C, U 6= C, ∃f : D → U where D denotes the open unit disk, which is a holomorphic bijection with holomorphic inverse (and essentially unique).
51
CHAPTER
14
FOURIER TRANSFORM AND SCHWARZ REFLECTION PRINCIPLE
We will go over contour integration via Fourier transforms and infinite sums, §11, the Schwarz reflection principle §7.2, and the Mobius transformations §13.2. The Fourier Transform. Let f : R → R or R → C. The Fourier transform of f is given by Z ∞ fˆ(y) = f (x)e−2πixy dx. −∞
The inversion formula is
∞
Z
fˆ(y)e2πiyx dy.
f (x) = −∞ 2πiyx
This is writing f in terms of e , and has applications in physics, probability theory (proving CLT via generating functions), etc. Look at Wikipedia for a summary of how useful it is. Example. f (x) =
1 1+x2 .
fˆ(y) =
Z
∞
1 e−2πiyx dx. 1 + x2
−∞
Assume y ≤ 0. Let C(R) = C1 (R) + C2 (R) denote the semicircle contour with C1 (R) denoting the part on the real axis and C2 (R) the semicircle part. Then consider Z 1 e−2πiyz dz = 2πi · Res(gy (z); i) 2 1 + z C1 (R)+C2 (R) where gy (z) is the integrand. With a simple bounding argument, we note that Z
Z
R
gy (z) = C1 (R)
−R
52
1 e−2πiyx dx. 1 + x2
2πy Also, Res(gy (z); i) = e 2i . Thus fˆ(y) = πe2πy for y ≤ 0. Now for y ≥ 0 we use the same −2πy contour reflected over the real axis. Then Res(gy (z); −i) = e −2i =⇒
Z
R
−R
1 e−2πiyx dx = 2πi · n(C, −i) · Res(gy (z); i) = πe−2πy . 1 + x2
So fˆ(y) = πe−2π|y| . The applications of Fourier transformations abound. Frankly Prof. Cotton-Clay hasn’t used it much, but natural scientists do. A very nice source for getting to know the complexanalytic details of Fourier transforms is Stein and Shakarchi. P∞ 2 Example. Verify that n=1 n12 = π6 . P∞ 2 We can do this by showing n=−∞,n6=0 n12 = π3 . This is left as an exercise to the reader (or see the textbook). Note that this is ζ(2), and the method that you use for computing this generalizes to ζ(2n), n ∈ Z+ ; this was first computed by Euler. Conformal mappings. These are functions that preserve angles. The Schwarz reflection principle is useful as a tool here. Theorem. (Schwarz reflection) Let H denote the upper half-plane, {z ∈ C : Im(z) > 0}. Suppose that f is holomorphic in a region D ⊆ H, that f is continuous on D ∪ L where L = ∂D ∪ R, and that f is real on L. Then ( f (z) z ∈ D ∪ L g(z) := f (z) z ∈ D is holomorphic on D ∪ L ∪ D. Proof. First, g is continuous because both parts agree on L. Second, g is holomorphic on D, and on D: lim
h→0
g(z + h) − g(z) f (z + h) − f (z) = lim = f 0 (z) h→0 h h
exists, so g is holomorphic on D. Then use Morera’s theorem to show that g(z) is holomorphic. Mobius transformations. This answers the question: what are the injective, holomorphic ˆ → C? ˆ maps C ˆ → C ˆ is holomorphic (meromorphic on C), then f is a rational Recall that if f : C P (z) function. Solving α = Q(z) , we have P (z) − αQ(z) = 0, and the LHS is a polynomial of degree max(deg P, deg Q). For f (z) to be injective, we need max deg = 1. So f (z) =
az + b , cz + d
53
where (a, b) and (c, d) are linearly independent; equivalently, ad − bc 6= 0. Maps of the form above are called Mobius transformations. It is obvious that the inverse f −1 is given by f −1 (w) =
−dw + b cw − a
with ad − bc 6= 0. Proposition. If f (z) =
az + b , cz + d
g(z) =
αz + β , γz + δ
then letting
A C
B D
=
a c
we have f (g(z)) =
b d
α γ
β δ
,
Az + B . Cz + D
Proof. This is just a simple check.
54
CHAPTER
15 ¨ MOBIUS TRANSFORMATIONS
Today we’ll talk about Mobius transformations §13.2, cross ratios, and automorphisms of D and H. Recall that we defined a Mobius transformation to be a function f (z) of the following form: az + b f (z) = . cz + d ˆ i.e. the holomorphic maps C ˆ →C ˆ with holomorphic These are the automorphisms of C, inverse. We showed that composition translates into matrix multiplication: fM1 (fM2 (z)) = FM1 M2 (z), where M ’s are the matrix forms of the transformation. These 2 × 2 invertible matrices with complex coefficients are known as the general linear group of order 2 over C: a b GL2 C = : a, b, c, d ∈ C, ad − bc 6= 0 . c d GL2 C has Lie group structure (it is an algebraic group endowed with manifold structure, e.g. the group operations are C ∞ ). Let M = {mobius transformations}. Then, modding out by complex scalars 6= 0, we have in fact that M=
GL2 C = P GL2 C, C×
where P GL2 C is the projective linear group of order 2 over C, which is also a Lie group. We mod out by scalar multiples of C because a b λa λb f =f , c d λc λd has no other coincidences; e.g. multiplication by a complex scalar gives the same transformation. We now claim that the generators for M are as follows: Generators for M: 55
(1) Dilations/rotations: z 7→ az. (2) Translations: z 7→ z + b. (3) Inversion: z 7→ z1 . Claim. These generate M. Proof. We get the map z 7→ az + b from dilation and translation. Suppose c 6= 0; we want a map of the form z 7→ az+b cz+d . Well, z 7→(1,2) cz + d 7→(3)
az + b 1 7→(1,2) . cz + d cz + d
Let’s show that we can achieve the latter map. We can write az + b 1 A + Bcz + Bd ad − bc 1 a =A +B = =− + , cz + d cz + d cz + d c cz + d c for which we want B = f (z) = ad z + db by (1,2).
a c
and A = b − ac d = − ad−bc c . Note that setting c = 0 we have
Corollary. {Mobius transformations} : {circles and lines} → {circles and lines}. Proof. All of the generators do.
Let’s examine the dimensions of the groups we have above. Note that GL2 C has 4 complex dimensions, or 8 real dimensions (by the obvious identification C ∼ = R2 ). This is because there are 4 degrees of freedom in choosing elements in the matrix; another way would be to show this is to note that GL2 C is open in Mat2 C under the continuous determinant map, the latter of which has complex dimension 4. Since we are eliminating one degree of freedom by imposing the condition det(A) 6= 0 for A ∈ M, we see that dimC M = 3, or dimR M = 6. Given this, we can say something about the action on the 3 points {∞, 0, 1}: Lemma. Given f ∈ M and f 6= id, f has at most 2 fixed points (points w ∈ C : f (w) = w) in {∞, 0, 1}. 2 Proof. Let c 6= 0, so that f (z) = az+b cz+d and setting f (z) = z we have az + b = cz + dz, a which maps ∞ 7→ c 6= ∞ because c 6= 0. For c = 0, we have ∞ 7→ ∞, and az + b = z has ≤ 1 solution. Proposition. There exists a unique Mobius transformation f (z) sending α1 , α2 , α3 to ∞, 0, 1 respectively, and f (z) is given by f (z) =
(z − α2 )(α3 − α1 ) . (z − α1 )(α3 − α2 )
Note that if any αj = ∞, we cross out both terms in which it appears. Proof. The existence of f (z) is trivial. Uniqueness follows thus: if we have two transformations f, g satisfying the above properties, we consder g ◦ f −1 , which sends ∞ 7→ ∞, 0 7→ 0, 1 7→ 1. So three points are fixed, which implies g ◦ f −1 (z) = z =⇒ f (z) = g(z). This suggests that dimC M = 3 is no mistake. Indeed, we can define the cross-ratio
56
[α1 : α2 : α3 : α4 ] as the number w ∈ C where α4 is sent if α1 7→ ∞, α2 7→ 0, and α3 7→ 1 by a Mobius transformation. Note that this number is equal to (α4 − α2 )(α3 − α1 ) . (α4 − α1 )(α3 − α2 ) Proposiiton. The cross-ratio is preserved by Mobius transformations, i.e. if f (z) = az+b cz+d with ad − bc 6= 0, then [α1 : α2 : α3 : α4 ] = [f (α1 ); f (α2 ); f (α3 ); f (α4 )]. Proof. Let g be a Mobius transformation sending α1 7→ ∞, α2 7→ 0, α3 7→ 1. We have [α1 ; α2 ; α3 ; α4 ] = g(α4 ). Then g ◦ f −1 sends f (α1 ) 7→ ∞, f (α2 ) 7→ 0, f (α3 ) 7→ 1, and g ◦ f −1 (f (α4 )) = g(α4 ). Proposition. There exists a non-unique Mobius transformation sending z1 7→ w1 , z2 7→ w2 and z3 7→ w3 , and it satisfies, for w = f (z), (z − z2 )(z3 − z1 ) (w − w2 )(w3 − w1 ) = . (w − w1 )(w3 − w2 ) (z − z1 )(z3 − z2 ) Proof. Existence comes from the formula and the cross-ratios. Uniqueness follows as before. Let’s use the above propositions above to find a Mobius transformation f sending D → H bijectively. If we visualize how f sends D ⊂ C → H ⊂ C, we can also visualize how it maps ˆ In particular, it should send −1 7→ ∞, 0 7→ i, 1 7→ 0, i 7→ 1, −1 7→ −1. Let w = f (z). C. Then using the equation above we have (z − 0)(1 + 1) 2z −z + 1 (w − i)(0 − ∞) = =⇒ iw + 1 = =⇒ w = i . (w − ∞)(0 − i) (z + 1)(1 − 0) z+1 z+1 Claim. This sends D to H. Proof. Let’s go with a more geometric proof. Note that f (i) = 1. We have a subclaim: ˆ there exists a unique circle or line through the 3 points. given 3 points in C, Proof of subclaim. Send the 3 points to ∞, 0, 1 by a Mobius transformation. There’s a unique circle/line through these, namely R. We showed that Mobius transformations sends circles and lines to circles and lines. Now f sends the unit circle to the real line by the formula f (eit ) = tan(t/2), where 0 ≤ t < 2π. We make another subclaim, namely that f sends points inside D to points inside H. Proof of subclaim. Use the following proposition on curves that are circles and omits −1 (a “Pac-Man” curve). Proposition. Let f (z) be a Mobius transformation, α ∈ C, γ a contour curve, and α ∈ / the interior of γ. Suppose n(f (γ); f (∞)) = 0. Then n(γ; α) = n(f (γ); f (α)). Proof. Under dilations, Z Z 1 0 1 dz 1 γ (t)dt n(γ; α) = = dt. 2πi γ z − α 2πi 0 γ(t) − α The equalities for the other generators follow similarly. In particular, note that for inversion we want to show the formula n( γ1 ; α1 ) = n(γ; α) + n( γ1 ; 0), from which we can conclude the 57
proposition. This is a simple check using the definition of winding number and the chain rule. Maps from regions into other regions. We know that log maps to a horizontal strip. In particular, log(z) : H → {z : 0 < Im(z) < π}, so 1 log(z) : H → R × [0, 1]. π We then see that D →f H →g R × [0, 1] for f (z) = i 1−z 1+z , g(z) = strip given by
1 π
log(z). Thus we have an automorphism from the unit disc to a 1−z 1 log(i ) : D → R × [0, 1]. π 1+z
58
CHAPTER
16 AUTOMORPHISMS OF D AND H
We were looking at H = {z ∈ C : Im(z) > 0} and D = {z ∈ C : |z| < 1}. −z+i Claim. f (z) = −i z−1 z+1 maps D to H, and g(z) = z+i maps H to D. Each are one-toone, holomorphic, and inverses of each other. Proof. Check the last part; we will show that f maps D to H. Note that, under f , 1 7→ 0, −1 7→ ∞, i 7→ −i( i−1 i+1 ) = 1. Hence, like last time, we see that the unit circle is mapped to the real line, since the 3 points determine a circle or a line. (Also note 0 7→ i, −i 7→ −1.) Apparently, the real axis is mapped to the imaginary axis. Now fill up D by circular arcs through −1 and 1. These map to straight lines “through 0 and ∞”, all ⊂ H because they intersect the upper half of the unit circle, which is the image of the imaginary axis ⊆ D. Another proof is the use the argument principle. Notice that the winding of f (γ(t)) around α ∈ C is the number of zeroes of f (z) = α... but this argument is sketchier, so we leave it to you to fill in the gaps.
Here’s an unnecessary but amusing claim: Claim. Let γ(t) be a contour curve and 0, α ∈ / im(γ). Then 1 1 1 , = n(γ, α) + n , 0 = n(γ, α) − n(γ, 0). n γ α γ Proof. We have Z n(1/γ, 0) = 1/γ
dz = z−0
Z 0
1
1 d( γt ) 1 γ(t)
−0
Z = 0
1
−γ 0 (t) dt = −n(γ, 0). γ(t)
On the other hand, we have Z 1 d( 1 ) Z 1 Z 1 Z 1 −γ 0 (t)/γ(t)2 −αγ 0 (t) αγ 0 (t) γ(t) n(1/γ, 1/α) = = = dt = dt 1 1 2 0 γ(t) − α 0 1/γ(t) − 1/α 0 αγ(t) − γ(t) 0 γ(t)[γ(t) − α] Z 1 Z 1 −γ 0 (t) γ 0 (t) = dt + dt = n(1/γ, 0) + n(γ, α), γ(t) 0 0 γ(t) − α 59
which gives the result.
Automorphisms of D and H. What are the injective, surjective, holomorphic maps D → D with holomorphic inverses? Any such map is in the automorphism group of D, denoted Aut(D). Recall that the Schwarz lemma tells us that if f : D → D is such that f (0) = 0, we have |f (0)| ≤ 1 and |f (z)| ≤ |z| for all z ∈ D, with equality in either where f (z) = eiθ z. Corollary. If f : D → D, f (0) = 0, and f is an automorphism of the disk (there exists a holomorphic inverse f −1 : D → D), then f (z) = eiθ z. Proof. |f (z)| ≤ |z| =⇒ |f −1 (z)| ≤ |z| so |z| ≤ |f (z)|, e.g. equality is obtained and f (z) = eiθ z. α−z for |α| < 1. This maps D → D and 0 7→ α, α 7→ 0. Hence Recall that Bα (z) = 1−αz Bα (Bα (z)) maps D → D, 0 7→ 0, α 7→ α, and their composition is the identity map.
Theorem. The automorphisms of D are precisely the maps α−z iθ f (z) = e . 1 − αz Proof. Suppose f is an automorphism of D with f (α) = 0. Then f (Bα (w)) takes 0 to 0 and f (Bα (w)) = eiθ w. Take w = Bα (z) and Bα (w) = z. Then f (z) = eiθ Bα (z) = α−z eiθ ( 1−αz ). Corollary. The automorphisms of D are the Mobius transformations taking D to itself. Corollary. The automorphisms of H are the Mobius transformations taking H to itself. Theorem. The automorphisms of H are precisely the maps f (z) = az+b cz+d with ad − bc > 0 and a, b, c, d ∈ R. Proof. Take x1 7→ ∞, x2 7→ 0, x3 7→ 1 in R ∪ {∞} via a Mobius transformation for x1 , x2 , x3 all distinct (note that there is a unique such Mobius transformation). Write it down: (z − x2 )(x3 − x1 ) . f (z) = (z − x1 )(x3 − x2 ) The collection of these is the collection of maps f (z) = f (i) ∈ H, f (i) =
(bd+ac)+i(ad−bc) . c2 +d2
az+b cz+d ,
ad − bc 6= 0. Now check when
We see that f maps into H iff ad − bc > 0.
Here is a classification of the automorphisms of D (and H), up to conjugation (e.g. for f, g : D → D, g −1 f g : D → D is allowed): • Identity: f (z) = z • Elliptic automorphisms: f (z) = eiθ z (rotation on D) • Parabolic: f (z) = z + b on H • Hyperbolic: f (z) = az on H, for a ∈ R Claim. Every automorphism of D (or H) is conjugate to precisely one of these. 60
Proof. Assume f is not the identity. We claim that every automorphism of H (or D) fixes either one point inside H, one point on R, two points on R, and no others. To see this, note that every Mobius transformation fixes one or two points. If one is in H, then the full Mobius transformation is given by the Schwarz reflection z∈H f (z) g(z) = something in R ∪ {∞} z ∈ R ∪ {∞} f (z) z ∈ H. So if we have one in H, then there can be no more. Either this happens, or one/two points are on R. If one fixed point is in H, we conjugate to D and see it’s at α ∈ D; conjugate by Bα to get it at 0, then get a rotation. In particular, let f : H → H, f (β) = f (β). Take our F : D → H, F −1 : H → D. F −1 ◦ f ◦ F then fixes F −1 (β) = α. Then taking Bα (which is its own inverse), we have Bα ◦ F −1 ◦ f ◦ F ◦ Bα sends 0 7→ 0, D → D, so it equals eiθ z. If we fix one point on R, we conjugate to the disk, then conjugate by rotation so that the fixed point on the unit circle is at −1. Then we conjugate back to H so that the fixed point is at ∞ (of Mobius transformation); i.e. f (z) = az + b. We need b 6= 0, else 0 7→ 0. We also have a > 0 by ad − bc > 0. If a 6= 0, we fix another point so that a = 1 and f (z) = z + b. If we fix two points in R, we again place one at ∞ and conjugate the other by A(z + 1) for appropriate A; then f (z) = az, a > 0. In fact, these are the symmetries of the hyperbolic plane (taking the line y = 0 to be a sort of “line at infinity”). As an aside, the hyperbolic metric on H is given by ds2 =
dx2 + dy 2 , y2
for y > 0. The arc length of γ(t) in the hyperbolic metric is γ(t) = (x(t), y(t)) : [a, b] → H.
61
R
ds = γ
R b √x0 (t)2 +y0 (t)2 a
y(t)
dt for
CHAPTER
17
SCHWARZ-CHRISTOFFEL AND INFINITE PRODUCTS
Today we’ll be talking about Schwarz-Christoffel functions §13.3 and infinite products and Euler’s formula for sin(πθ), §17.3. Schwarz-Christoffel maps. These are holomorphic maps from H to a single polygon. We’ll see next time how it follows from the Riemann mapping theorem. But first off we can note that there are only a limited number of ways to do this. Let’s draw a polygon with vertices vi , interior angles πβP i , and exterior angles παi P (παi + πβi = π). Note that −1 < αi < 1, and παi = 2π =⇒ αi = 2. Proposition. Given angles πα1 , ..., παn and that a1 , ..., an−1 are points on R, then Z z dz f (z) = α αn −1 1 0 (z − a1 ) ...(z − an−1 ) gives a holomorphic map from H to a polygon with angles αi , and has a continuous extension to R with f (ai ) = vi , f (∞) = an . Remarks on the function. This is a contour integral taken on H ∪ R. We can take a branch of (z−a1j )αj = (z − aj )−αj which is positive for z real, > aj and extends downward, e.g. is defined on C − {aj − ir}, r > 0. We can require 0 6= aj , but in fact this still makes R1 sense if 0 is one of those: 0 x1λ dx converges for λ < 1. Let’s examine the argument of Qn−1 f 0 (z) = j=1 (z − aj )−αj on R: Claim. Arg(f 0 (z)) is constant on the connected segments of R − ∪{αj }. Also, Arg(f 0 (z)) increases by παj when preserving aj . Proof. This follows from: Claim. Let z −α be defined on C − {negative imaginary axis}, and be and positive real for z real and positive. Then Arg(x−α ) = 0 for x > 0. If x is real, Arg(x−α ) = −πα, x < 0. Proof. Write z = reiθ . Then z −α = r−α e−iαθ gives the desired branch, which is in particular our negative real axis. So if x is negative, i.e. λ = reiπ , then we get x−α = r−α e−iαπ , i.e. Arg(x−α ) = −απ. We can conclude that on the boundary (R) of H, f (z) maps to 62
straight lines, turning by angles παk to L.
R∞ Q Claim. This closes up, e.g. −∞ (x − aj )αj dx = 0. Proof. Use a semicircle contour, but skip over the poles on the real line. Then use a combination of the closed curve theorem and bounding on the arc. Example. sin z maps (− π2 , π2 ) × (0, ∞) to H. We can view the domain as a triangle with exterior angles π/2, π/2, π. So, in some sense, sin−1 z : H → this strip. We can then write Z z Z z dz dz √ = , sin−1 (z) = 1/2 (1 − z)1/2 2 (1 + z) 1 − z 0 0 for some branch of the root function. (This is because
−1 d dx (sin
x) =
√ 1 .) 1−x2
Infinite Q∞ products. We want to segue into the question, when does the infinite product k=1 ak converge? Q P Lemma. ak converges log(aP k ) converges, for Re(ak ) > 0. Qn iff n Proof. Let pn := k=1 ak , sn = k=1 log(ak ). Then log pn = sn and pn = esn . e and log are continuous, so convergence for one gives convergence for another. Q P Lemma. (1 |ak | converges. P+ ak ) converges if P∞ Proof. If |ak | converges, then past some point N , |ak | < 1/z. So we check: k=N log(ak ) converges. 1 1 a3 a4 a2 | log(1 + ak )| = ak − k + k − k + ... ≤ |ak | 1 + + + ... ≤ 2|ak |. 2 3 4 2 4 P∞ P P∞ So N | log(1 ak )| ≤ 2|ak | converges, hence N log(1 + ak ) converges; finally, this Q+ ∞ implies that N (1 + ak ) converges. P Proposition. Suppose fk (z) is holomorphic on a region D, with |fk (z)| uniformly converQ gent on compact subsets of D. Then k=1 (1 + fk (z)) is uniformly convergence on compact subsets and hence convergent. Proof. Use the proof of the lemma in uniform convergence along with Morera’s theorem. Example. (Euler, 1734) We have the infinite product formula for sin(πz): ∞ Y z2 sin(πz) = πz 1− 2 . k k=1
Theorem. (Weierstrass product) There exists an entire, holomorphic function with zeroes only at {αk }k=1 if ak → ∞. Furthermore, the order of zero is the number of times repeated in the list. Q Q Proof. First attempt to go about it: f (z) = (z − ak ) =⇒ f (z) = (1 − azkk ) converges P 1 if |ak | converges.
63
Claim. If
P
1 |ak |2
converges, then f (z) =
Y z 1− ez/ak . ak
P Proof. This converges if |(log(1 − azk ) + azk )| does. Suppose is possible for k > N since ak → ∞. Then we can bound |z|2 1 1 |z|2 ≤ 2 1 + + + ... ≤ 2 , ak 2 4 |ak |2 P 1 i.e. suppose |ak |2 converges.
|ak | 2
> |z| for |z| ≤ R; this
Back to the full proof of Weierstrass’s theorem: Let Y z f (z) = 1− Ek (z) , ak for Ek (z) = e
z ak
2
z + 2a +...+ k
zk kak k
.
Check via a diagonal argument that this works; in particular, consult your textbook.
Proposition. (Euler’s formula) sin πz = πz
Y
z2 1− 2 k
Proof. Take the quotient πz
Q 1−
Q(z) =
z2 k2
. sin πz This is an entire function with no zeroes. We’ll argue that this has growth at most Aez ; let’s claim that it follows that Q(z) is constant. R z 0 (z) We note that Q(z) is even. So, taking the log of Q(z), we have 0 QQ(z) dz. This has growth at most constant = |z|3/2 for large values. So log Q(z) is linear, i.e. Q(z) = AeBz with B = 0 since Q is even, i.e. Q(z) is constant. Q(z) is constant because limz→0 sin(πz) =1 πz Q∞ z2 and limz→0 k=1 (1 − k2 ) = 1. Now let’s talk about the growth. sin1πz is bounded on a square with sides from −N − 1/2 to N + 1/2 (exercise), and likewise on the edges. To bound the rest...actually, we’ll save that for next time because we’re out of time.
64
CHAPTER
18 RIEMANN MAPPING THEOREM
Guest lecture: Joe Rabinoff Note: These notes are particularly messy since the scribe wasn’t paying too much attention, so use in discretion. You’re lucky that we cover the Riemann mapping theorem, which is one of the most interesting results you’ll learn in this class. The context is the following: ˆ −R Let regions R1 , R2 ( C be open and simply connected, with R1 connected and C connected. Then there exists ϕ : R1 → R2 that is onto, one-to-one, and holomorphic. Thus ϕ−1 : R2 → R1 is also holomorphic. As far as complex analysis is concerned, these regions are “identically the same.” Now let’s introduce a bit of reductionism here. Assume ψ : R2 → U = {x ∈ C : |x| < 1}, so that we have the following diagram: ϕ R1 R2 ϕi
ψi
U We need R1 6= C: if ϕ : C → U , then ϕ is bounded, then constant by Liouville’s theorem. Theorem. Let R ( C be open, R 6= ∅, and R 6= C. Choose a z0 ∈ R. Then there exists a unique, one-to-one, holomorphic map ϕ : R → U such that ϕ(z0 ) = 0 with ϕ0 (z0 ) ∈ R≥0 . Proof. Uniqueness: suppose ϕ1 , ϕ2 : R → U both satisfy the theorem. Then ϕ : ϕ2 ◦ϕ−1 1 : U → U is one-to-one, onto, and holomorphic, and ϕ(0) = ϕ2 (ϕ−1 (0)) = ϕ2 (z0 ) = 0. By the Schwarz lemma, we know that |ϕ(z)| ≤ |z| for all z ∈ U , and if ϕ is onto, ϕ(z) = eiθ z. Then ϕ−1 (0) = eiθ ∈ R≥0 =⇒ eiθ = 1 =⇒ ϕ(z) = z. This implies ϕ0 (0) ϕ2 = ϕ−1 = Id =⇒ ϕ2 = ϕ1 , as ϕ0 (0) = ϕ02 (0) > 0. So uniqueness is easy. Let’s work 1 1 out an example for the other parts: Example. Let R = U , ϕ : R = U → U . If ϕ(0) = 0, then by Schwarz’s lemma, we
65
know that ϕ is onto ⇐⇒ |ϕ0 (0)| = 1 =⇒ |ϕ0 (0)| is the largest possible among all maps U → U if 0 7→ 0. α−z is such that Bα : U → U , Bα (0) = α, If ϕ : U → U, ϕ(0) = α 6= 0, then Bα (z) = 1−αz Bα (α) = 0; e.g. Bα swaps 0 and α. Then |(ϕ ◦ Bα )0 (0)| = |ϕ0 (α)||Bα0 (0)| ⇐⇒ |ϕ0 (α)| is constant. It turns out that ϕ is one-to-one and onto =⇒ |ϕ0 (α)| is maximized. In general, we want to find ϕ : R → U that’s one-to-one and holomorphic, with ϕ(z0 ) = 0, |ϕ0 (z0 )| maximized. Even more precisely, we will take F = {f : R → U : f is one-to-one, holomorphic, and f 0 (z0 ) = 0}. We’ll show (a) F 6= ∅ (b) supf ∈F f 0 (z0 ) = M < ∞ (b’) There exists f ∈ F : f 0 (z0 ) = M . (c) If ϕ = f from above, then ϕ : U → U is one-to-one and onto, ϕ(z0 ) = 0, and ϕ0 (z0 ) > 0. In this case, the function we will cook up looks like Bα . First we want f : R → U that is one-to-one and holomorphic. Suppose there exists δ . Then this D(p0 , δ) := {z : |z − p0 | < δ} such that D(p0 , δ) ∩ R − ∅. Let f (z) = z−p 0 δ δ is one-to-one since it’s a Mobius transform, if z ∈ R =⇒ |f (z)| = |z−p0 | < δ = 1. q z−p0 What is no such disc exists? Choose any p0 ∈ / R, and define g(z) = z0 −p0 with g(Z0 ) = 1, one-to-one since g(z)2 is. Do this by choosing the appropriate square root. How √ do we choose a branch of ? Well, r
and
log
z − p0 z0 − p0
z − p0 = exp z0 − p0
1 log 2
z − p0 z0 − p0
Z
z
= log(z − p0 ) − log(z0 − p0 ) = z0
This makes sense by the closed curve theorem, as
1 ζ−p0
dζ . ζ − p0
is holomorphic on R.
Claim. There exists δ > 0 such that |g(z) + 1| > δ for all z ∈ R. Proof. If not, then ∃z0 , z1 , ... ∈ R such that g(zi ) → −1. Then
q
zi −p0 z0 −p0
→ −1, and
zi −p0 z0 −p0
squaring = 1 =⇒ lim(zi − p0 ) = lim(z0 − p0 ) =⇒ lim zi = z0 . But by continuity of g, lim g(zi ) = g(z0 ) = 1, a contradiction. This proves (a). Proof of (b). Since R is open, D(z0 , 2δ) ⊆ R for some δ > 0. If f ∈ F, then Z Z 1 Z 0 f (z) |f 0 (z)| 1 1 1 1 0 |f (z0 ) = dz ≤ dz ≤ dz = 2 2 2πi C(z0 ,δ) (z − z0 )2 2π |z − z0 | 2π δ δ =⇒ sup |f 0 (z)| ≤ f ∈F
66
1 0, and Sj := {z ∈ K : |ϕn (z) − ϕ(z)| < for all n > j}. Sj is open: indeed, if z1 ∈ Sj and |z1 − z2 | < dδ for δ > 0, then |ϕn (z1 ) − ϕ(z2 )| ≤ |ϕn (z1 ) − ϕn (z1 )| + ... ≤ |ϕn (z1 ) − ϕ(z)| + 2δ < /2 for all n > j. Taking δ small enough, z2 ∈ Sj . This implies D(z, δ) ⊆ Sj . ∞ [
Sj = K
j=1
Compactness implies K + Sj for some j implies uniform convergence. All you do now is fiddle around with Bα ’s.
68
CHAPTER
19 RIEMANN MAPPING THEOREM AND INFINITE PRODUCTS
We will continue with the Riemann mapping theorem, which we got through most of last time: ˆ Theorem. (Riemann mapping) Let R ⊂ C be open and simply connected (for us, use C−R is connected, i.e. can apply closed curve theorem). Also suppose R 6= C. Then ∃! map f that is holomorphic, with holomorphic inverse, s.t. f : R → D, with f (α) = 0, R≥0 3 f 0 (α) > 0. Improperly Stated Generalization. (Uniformization theorem) Every simply connected Riemann surface is conformal to, and has a holomorphic map with holomorphic inverse to, precisely one of ˆ = S2 (1) C (2) C (3) D The nice thing is that these correspond to different geometries. D is the hyperbolic geometry, S 2 you’ve done on homework, and C is Euclidean. We don’t have the tools to do this properly yet. Proof of the Riemann mapping theorem so far. Let F = {f : R → D : f is holomorphic, 1-1, and f 0 (α) > 0}. We show that (A) F 6= ∅. (B) supf ∈F |f 0 (α)| < ∞ and ∃g ∈ F achieving this sup, e.g. |g 0 (α)| = supf ∈F |f 0 (α)|. (C) Given this g achieving the sup, g is surjective and g(w) 6= 0 for w ∈ R (so that we get a holomorphic inverse). We can also draw pictures for A and B, which we showed last time with the help of the following: Lemma. (Montel’s theorem) A uniformly bounded sequence of holomorphic functions has a subsequence which converges on compact subsets of the domain, and the limit is 69
holomorphic (using Morera’s theorem). For more info here, see also the Arzela-Ascoli theorem. Let’s finish up with C. Claim. This limit g has the property that g(α) = 0 and is surjective (any 1-1 function f can have f 0 (w) = 0 for any w. ). Proof. g sends 0 7→ 0. If not, g(α) = w 6= 0, and Bh ◦ g : α 7→ 0. Then |(Bw ◦ g)0 (α)| = 1 0 0 |Bw (g(α))g 0 (α)| = 1−|w| 2 |g (α)| > 1, and there is a larger such g. g is surjective. Suppose not, that that it misses w. Wlog w = 0, and consider (Bw ◦ g), 0 which sends α 7→ w and misses 0. Then |(Bw ◦g)0 (α)| = |Bw (g(α))||g 0 (α)| = (1−|w|2 )|g 0 (α)|. Now consider the following claim: Claim. ∃ a holomorphic branch of Proof. We can write
√
Bw ◦ g (which maps R → D) Z
z
log(Bw ◦ g)(z) − log(Bw ◦ g)(α) = α
Then take e
1 2 (∗)
(Bw ◦ g)0 (z) dz = ∗. (Bw ◦ g)(z)
.
√ √ √ √ We have ( ◦ Bw ◦ g) : w 7→ w by construction. Then ( ◦ Bw ◦ g)0 (α)| = | 0 (w)|(1 − √ √ 1 2 0 2 0 |w| )|g (α)| = 2|w|1/2 (1 − |w| )|g (α)|. Also, (B√w ◦ ◦ Bw ◦ g) : α 7→ 0, and (B√w ◦ ◦ √ 1 1 0 Bw ◦ g)0 (α) = |B√ ( w)|... = 1−|w| (1 − |w|2 )|g 0 (α)|. w 2|w|1/2 Set |w| = r2 , for r ∈ R≥0 (with w not 0 b/c g 0 (α) > 0). We claim that 1 1 (1 − r4 ) > 1 2 1−r 2r for 0 < r < 1. To see this, note that 1 + r2 − 2r = (r − 1)2 > 0 for 0 < r < 1, so
1+r 2 2r
> 1.
So g is surjective. Hence g is injective. We claim that g 0 (w) 6= 0 for any w. To see this, suppose g 0 (w) = 0. Wlog suppose w = 0. Then g 0 (0) = 0 and g(z) = z k h(z) for k ≥ 2, with h holomorphic and h0 (0) 6= 0. Hence Z g 0 (z) dz = Z(g) C (0) g(z) inside D (0) = k. Take β with |β| small; then Z g 0 (z) k= dz = Z(g − β) C (0) g(z) − β in D (0). If so, g(z) − β is 0 at inverse images of β, but all of the same small ball around 0 is an inverse image of something. So g 0 (0) = 0 and g is constant. 70
Here are some consequences of the Riemann mapping theorem: (1) Every connected and simply connected region in R2 is topologically equivalent (in the sense of homeomorphic) to a disk. For example, consider taking out a Cantor set in R2 . (2) See moduli spaces of curves and their relations to string theory. Back to infinite products and the gamma function. Let’s finish the proof of the following theorem: Theorem.
∞ Y z2 sin(πz) = πz 1− 2 . k k=1
Corollary. Set z = 1/2, and 1=
∞ ∞ π Y 1 π Y (2k − 1)(2k + 1) 10 = . 2 (2k)2 2 (2k)(2k) k=1
Then
k=1
π = 2
2·2 1·3
4·4 3·5
6·6 5·7
....
This has a taste of number theory. Proof of the theorem. The gist of the proof is to write Q(z) =
πz
Q∞
− sin(πz) k=1 (1
z2 k2 )
, 3/2
which has growth order of at most something like Ae|z| . We conclude that Q(z) is a constant, which taking limits we see to be = 1. See the textbook for a more thorough proof. This is presumably the flavor of analytic number theory, where we can’t show things definitely but can use complex analysis to give a (poor) bound that translates into something nice. We’ll continue with the gamma function Γ(z) next time.
71
CHAPTER
20
ANALYTIC CONTINUATION OF GAMMA AND ZETA
Analytic continuation of gamma and zeta. So we have these two functions: Z ∞ ∞ X 1 Γ(z) = e−t tz−1 dt, ζ(z) = z n 0 n=1 The first is holomorphic for Re(z) > 0 and the second for Re(z) > 1; this is just an application of Morera’s theorem. We’ll talk about the gamma function for a while; what we’d like is for these functions to be holomorphic in all of the complex plane. It turns out that the properties of the gamma function will be important for understanding prime numbers. Let’s start with gamma—recall what happened last time. We have Z ∞ Z 1 Z ∞ Γ(z) = e−t tz−1 dt = e−t tz−1 dt + e−t tz−1 dt. 0
0
1
The right integral uniformly converges on BR (0), i.e. it converges on compact subsets of C. Now look at the left integral. We saw that it converges uniformly for compact subsets of Re(z − 1) > −1, i.e. Re(z) > 0. How do we extend this to all of C? The extension will have various poles, and is meromorphic in C. Claim. Γ(z + 1) = zΓ(z). Proof. We integrate by parts. Consider Z Γ(z) =
∞
e−t tz−1 dt.
0
Take u = tz−1 , dv = e−t dt =⇒ du = (z − 1)tz−2 dt, v = −e−t . Then Z ∞ Z ∞ ∞ e−t tz−1 dt = −e−t tz−1 0 + (z − 1)e−t tz−2 dt = (z − 1)Γ(z − 1), 0
0
as desired.
Corollary. Γ(n) = (n − 1)! for n ∈ Z≥1 . 72
Proof. Check Γ(1) =
R∞ 0
e−t dt = 1 from the claim.
From the above, we can see that Γ(z + 1) z is meromorphic (holomorphic) for Re(z + 1) > 0, i.e. Re(z) > −1 with a pole only at z = 0. Repeating this k times, we see that Γ(z) =
Γ(z) =
Γ(z + k + 1) z(z + 1)...(z + k)
is meromorphic for Re(z + k + 1) > 0, i.e. Re(z) > −k + 1, with poles only at nonnegative integers. This gives a continuation of Γ into the left side of C. The residues at the poles is given by Res(Γ(z); 0) = Res(Γ(z); −k) =
Γ(1) = 1, 1
Γ(1) (−1)k = . (−k)(−k + 1)...(−1) k!
These poles are indeed simple by our calculations above. We will show another nice claim: Claim. Γ(z)Γ(1 − z) =
π sin(πz) .
Corollary. Γ has no zeroes. Proof. This property follows directly from the claim. sin(πz) is nonzero for z not an integer; for z an integer, we already know that Γ has a pole if z ∈ Z≤0 or Γ(z) = (z − 1)! 6= 0 if z ∈ Z≥1 . To prove the initial claim, we’ll need to show another fact, that for n t −t e = lim 1 − n→∞ n we have Z n→∞
n
1−
Γ(z) = lim
0
t n
n
tz−1 dt,
R e.g. we can switch the lim and . Then n Z n Z n t 1 lim 1− tz−1 dt = lim n (n − t)n tz−1 dt, n→∞ 0 n→∞ n n 0 and from integration by parts we have n Z n z Z n tz t n z−1 n (n − t) t dt = (n − t) + n(n − t)n−1 dt. z z 0 0 0 So
1 n→∞ nn
Z
lim
n
1 n n→∞ nn z
(n − t)n tz−1 dt = lim
0
73
Z 0
n
tz (n − t)n−1
Z n 1 n(n − 1)...1 tz+n−1 dt n→∞ n2 z(z + 1)...(z + n − 1) 0 n z n nn nz 1 n tz+n = lim = ... since . n→∞ z z+1 z+n z + n 0 z+n = lim
This gives z 1 = lim Γ(z) n→∞ nz
z+1 1
...
z+n n
n z z Y 1 + , n→∞ nz k
= lim
k=1
which looks something like the expression for sin(πz) we saw before. Let’s rewrite the last expression as n ∞ z Y z z z Y h z −z/k i z −z 1 + = lim zn + + ... + 1+ e , exp z + z n→∞ n n→∞ k 2 3 n k
lim
k=1
k=1
and the product by itself converges as shown before. Rewriting once again with n−z = e−(log n)z , ∞ z z z Y h z −z/k i lim zn−z exp z + + + ... + 1+ e n→∞ 2 3 n k k=1 " !# ∞ n X Y h 1 z −z/k i = lim exp z − log n z 1+ e . n→∞ k k k=1
k=1
Also, " lim
n→∞
n X 1
k=1
k
# − log n = γ ≈ 0.577,
the Euler-Mascheroni constant. Then ∞ h Y z −z/k i 1 = e−γz z 1+ e . Γ(z) k k=1
From this we see that ∞ ∞ Y Y 1 1 −γz γz z2 z2 sin(πz) 1 = = e e z(−z) 1− 2 =z 1− 2 = , Γ(z)Γ(1 − z) −zΓ(z)Γ(−z) −z k k π k=1
which is a cute equation. It also gives the values √ √ 1 3 π “ = π, = Γ (1/2)!” = Γ . 2 2 2
The zeta function. We turn to ζ(z), defined as ζ(z) =
∞ X 1 , nz n=1
74
Re(z) > 1.
k=1
We claim the following: Claim. ζ(z) has a meromorphic extension to all of C with a simple pole only at 1. We’ll show next time that ζ(z) has no zeroes on Re(z) = 1, and from this deduce the prime number theorem, which relates to the growth rate of the number of prime numbers. You may have also heard of the Riemann hypothesis, which says that the nontrivial zeros of ζ are all on the line Re(z) = 1/2; this gives us a stronger estimate of the growth rate of the primes. Claim. For Re(z) > 1, ζ(z) =
Y p prime
1 1 − p−z
.
Proof. We first write 1 1 1 1 = 1 + z + 2z + 3z + ... 1 − 1/pz p p p αk 1 This gives us something of the form 1 + 1/2z + 1/3z + .... If we write n = pα 1 ...pk , then
1 1 1 = α1 ... αk nz p1 z pk z in the product. Then N Y X 1 1 1 1 + z + ... + M z ≤ z n p p n=1 p 1, and by symmetry there are no zeroes with x < 1. Note that Γ(z) has simple poles at nonnegative integers, while Γ(z/2) has simple poles at 0, −2, −4, .... Then ζ(z) =
π z/2 ξ(z) Γ(z/2)
has trivial zeroes at −2, −4, ..., and we’ll show that there are no zeroes right of x = 1 (also, there is neither a pole nor zero at 0). Claim. ζ(z) has no Q zeroes in {Re(z) > 1}. Proof. ζ(z) = p prime (1 − 1/pz ) ≥ 1 for Re(z) > 1.
Theorem. ζ(z) 6= 0 for Re(z) = 1. Proof. Deferred to below. Theorem. (Prime Number) Let π(N ) = # of primes ≤ N . Then π(N ) ∼ i.e.
N , log N
π(N ) →1 N/ log N
as
N → ∞.
We won’t show this theorem today, but we will show that ζ(z) has no zeroes for Re(z) = 1. 80
To do this we need a few other results: Lemma. X
log ζ(z) =
m∈Z,p prime
1 . mpmz
Proof. Y
ζ(z) =
p prime
log ζ(z) = −
X p prime
1 log 1 − z p
1 1 − 1/pz
∞ X X (p−z )m = = m m=1 p prime
X m∈Z,p prime
1 . mpmz
Note that this is much like the zeta function: ∞ X X (p−z )m = m m=1
p prime
where
X m∈Z,p prime
∞ X cn 1 = , z mpmz n n=1
( 1/m n = pm cn = 0 otherwise
and cn ≥ 0. This is an instance of a Dirichlet series, which we won’t talk about in detail. (Take Math 229x for this.) Lemma. 3 + 4 cos θ + cos 2θ ≥ 0. Proof. Trivial.
Lemma. For z = x + iy, x > 1, log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 0. Proof. We have log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = 3 log |ζ(x)| + 4 log |ζ(x + iy)| + log |ζ(x + 2iy)| = 3Re(log ζ(x)) + 4Re(log ζ(x + iy)) + Re(ζ(x + 2iy)) X = n−x (3 + 4 cos(y log n) + cos(2y log n)) ≥ 0, n
as desired.
Proof of the theorem. We wish to show that ζ(z) 6= 0 for Re(z) = 1. Suppose ζ(x + 2iy) = 0. Then consider ζ(x)3 ζ(x + iy)4 ζ(x + 2iy) = A. If ζ(1 + iy) = 0, then as x → 1+ , A → 0, a contradiction. This is because ζ(z) has a simple pole at z = 1, ζ(z) has zero at z = 1 + iy, and ζ(z) is holomorphic (no pole) at z = 1 + 2iy. Chebyshev/Tchebychev’s ψ(x) function. (unrelated to ψ(u) earlier) Define X X X log x ψ(x) := log p = Λ(n) = log p, log p pm δ > 0. • Fact 2: If k is odd, then Ek (τ ) = 0. • Fact 3: We have the following transformation relations: Ek (τ + 1) = Ek (τ )
and
Ek (−1/τ ) = τ k Ek (τ )
E.g. for the last relation we have (m + n(−1/τ ))k = τ −k (mτ − n)k . These properties tell us that E2k is a weakly modular function of weight 2k. Note that, since the Eisenstein series converge depending on τ , we can just write them as constants. Proposition. If z is near 0, then ∞
℘(z) =
X 1 1 + (2k + 1)E2k+2 z 2k = 2 + 3E4 z 2 + 5E6 z 4 + ... 2 z z k=1
Proof. We can write X 1 1 1 ℘(z) = 2 + − 2 , z (z − w)2 w ∗ w∈Λ
where we reindex the lattice under the transformation w 7→ −w. Then 1 1 1 = 2 (z − w)2 w (1 − z/w)2 1 1 ∂ 1 2 Noting that 1−x = 1 + x + x2 + ... for |x| < 1 and (1−x) 2 = ∂x ( 1−x ) = 1 + 2x + 3x + ..., we have for |z/w| < 1 that z 2 1 1 z = 1 + 2 + 3 + ... . (z − w)2 w2 w w
91
So
∞ ∞ X 1 X X 1 1 zn ℘(z) = 2 + = + (n + 1) z w2 n=1 wn z 2 n=1 w∈Λ∗ P 1 Since En+2 = w∈Λ∗ wn+2 , we have
X w∈Λ∗
1 wn+2
! (n + 1)z n
∞
℘(z) =
X 1 + (2k + 1)E2k+2 z 2k , 2 z k=1
as En+2 = 0 when n is odd.
Theorem. If ℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 , then g2 = 60E4 and g3 = 140Eg . Proof. We can write out some terms of the identity we derived above: ℘(z) =
1 + 3E4 z 2 + 5E6 z 4 + ... z2
2 + 6E4 z + 20E6 z 3 + ... z3 4 1 ℘0 (z)2 = 4℘(z)3 + ... = 6 − 24E4 2 − 80E6 + ... z z 1 1 ℘(z)3 = 6 + 9E4 2 + 15E6 + .... z z Now take the difference ℘0 (z) = −
℘0 (z)2 − 4℘(z)3 + 60E4 ℘(z) + 140E6 . This is holomorphic near 0; each term is holomorphic away from Λ and is doubly-periodic, so the difference is holomorphic, doubly-periodic, and takes 0 to 0. Then, from earlier problems, we know that it’s 0, and the equality ℘0 (z)2 = 4℘(z)3 − g2 ℘(z) − g3 gives g2 = 60E4 and g3 = 140Eg as desired.
What is this explicit formula telling us? On the one hand, we have C/Λ, the “domain of ℘ and ℘0 ” which we identify with a torus. Consider a map to C2 given by z 7→ (℘(z), ℘0 (z)); we can draw C2 (which we don’t really discuss in class) by drawing R × R on the axes in 2-space, and the curve it traces out is the set of solutions to the polynomial function y 2 = 4x3 − g2 x − g3 , a cubic curve. What you’re seeing is that this complex torus is a cubic curve in C2 . We can check injectivity and surjectivity and other nice things, but let’s get out of this story portion and move into.... The space of elliptic curves. Now we want to consider all possible τ ’s, and we note that the lattice Λ doesn’t change under the action of P SL2 Z ⊆ Aut(H) = P SL2 R. This is just the space of transformations of the form τ 7→
aτ + b cτ + d 92
for a, b, c, d ∈ Z, generated by τ 7→ τ + 1 and τ 7→ −1/τ ; it’s what you can do to τ to not change your lattice. (The elliptic functions compose with dilations; e.g. they’re maps from coplex tori from one direction to the other direction, just by dilation. You can’t just stretch it in one direction, since then it won’t be holomorphic.) So let me show you a fundamental domain D in H. This is the part of H above the circle |z| = 1 and bounded by the lines Re(z) = {−1/2, 1/2}; e.g. D = {z ∈ H : |z| > 1, − 21 < Re(z) < 12 }. This is the moduli space of complex 1-tori (elliptic curves). In other words, D is the fundamental domain of the action of P SL2 Z. H/P SL2 Z is the moduli space of elliptic curves. Let’s do some justice to why these things are nice. E2k (τ ) is sort of a function on H/P SL2 Z, defined on D, and transforms under the involution τ 7→ −1/τ : E2k (−1/τ ) = τ 2k E2k (τ ). When we say these things are called modular forms, we mean that they “don’t change the correct way,” and the reason here is the power coefficient. We can, however, write lim Im(τ )→∞
E2k (τ ) =
lim Im(τ )→∞
X (m,n)6=(0,0)
∞ X 1 1 = 2 = 2ζ(2k), 2k (m + nτ )2k m m=1
since for n 6= 0 the center experssion has the nτ term → 0 for Im(τ ) → ∞. We also have E4 (∞) = 2ζ(2k). Then
4 π2 ) = π4 . 90 3 8 6 g3 (∞) = 140E6 (∞) = 140(2)(ζ(6)) = π . 27 Setting the modular discriminant ∆ = g23 − 27g32 , we note that ∆(∞) = 0; the right hand side expression, g23 − 27g32 , is the discriminant of 4x3 − g2 x − g3 . This is a weight 12 modular form, by which we mean that E2k is a transformation of weight 2k, and it’s modular if limIm(τ )→∞ E2k (τ ) exists and holomorphic at ∞, E2k (τ ) is holomorphic on H, and E2k (τ + 1) = E2k (τ ). We have another modular form of weight 12, which is simply g23 . Define g2 (∞) = 60E4 (∞) = 60(2)(
j=
1728g23 . ∆
This has weight 0, so it’s really defined as a function on H/P SL2 Z (e.g. a meromorphic function from H → C invariant under the action of SL2 Z), which is the moduli space of all elliptic curves. It’s called a j-invariant of elliptic curves. There’s a theorem that states: Theorem. This is holomorphic on H, invertible under P SL2 Z, has j(∞) = ∞, with residue 1 at ∞, and gives a surjection H → C that injects from the quotient H/P SL2 Z. Proof. See Serre’s Course in Arithmetic. I won’t say too much more about this subject; we’ve got j as an invariant in the moduli space, but we won’t talk about why number theorists and cryptographers are interested in it. For this, consult any reference on elliptic curves. We’ll use the remainder of the time for questions. 93
APPENDIX
A
SOME THINGS TO REMEMBER FOR THE MIDTERM
Complex expressions: 1. Remember polar form: z = reiθ . u v 2. Stereographic projection is given by ϕ : S 2 −{p} → R2 , ϕ : (u, v, w) 7→ ( 1−w , 1−w ), ϕ−1 : 2
2
2y x +y −1 2x (x, y) 7→ ( x2 +y 2 +1 , x2 +y 2 +1 , x2 +y 2 +1 ). iz
−iz
iz
3. cos z = e +e , sin z = e −e 2 2i 4. log z = log r + iθ + i(2πn).
−iz
.
Holomorphic functions: 1. To show that a function f is holomorphic, use the Cauchy-Riemann equations and show that the partials are continuous. You can also any rule from real analysis (the chain rule, the product rule, the quotient rule, etc.), show that f has a power series development, or use Morera’s theorem. 2. Recall that the Cauchy-Riemann equations are ( ∂v ∂u = ∂y ∂x ∂v ∂u − ∂x = ∂y . R 3. Morera’s theorem: If f (z) is continuous on a region D and Γ f (z)dz = 0 for Γ a triangle (or closed curve), then f (z) is holomorphic. This is useful for: 1. Showing limn→∞ fn = f is holomorphic. 2. Showing some functions defined by sums or integrals are holomorphic. 3. Showing functions that are continuous and holomorphic on all but some set (e.g. some point) are holomorphic. 4. f is holomorphic ⇐⇒ f complex analytic ⇐⇒ f smooth; this means you can substitute any holomorphic function with its power series, and can differentiate it endlessly. 5. The closed curve theorem tells us that the integral of a function that is holomorphic in the open disk D over a contour C ⊂ D is 0. 94
6. Per the antiderivative theorem, if f (z) is holomorphic on an open disk D, then it has an antiderivative F (z) such that F 0 (z) = f (z). 7. Cauchy’s integral formula allows us to determine the value of a holomorphic function f at a point. In particular, Z 1 f (z) f (w) = dz, 2πi C z − w where f is holomorphic in an open disk D and C contains w. 8. The uniqueness theorem tells us that holomorphic functions are determined uniquely up to their values in a certain region. 9. The mean value theorem allows us to evaluate a holomorphic function at point. If f (z) is holomorphic on Dα (R), then ∀r : 0 < r < R, we have Z 2π 1 f (α + reiθ )dθ. f (α) = 2π 0 10. The maximum modulus principle: if f is holomorphic in any region, then its maximum is achieved at the boundary. 11. Harmonic functions are characterized by the fact that their Laplacian is zero, and possess nice properties as you discovered in the problem sets. Entire and meromorphic functions: 1. Entire functions are functions that are holomorphic in all of C. 2. Liouville’s theorem tells us that any bounded entire function is constant. A generalization is that if |f (z)| ≤ A + B|z|N and f (z) is entire, then f (z) is a polynomial of degree ≤ N . 3. Likewise, if f (z) is entire and f (z) → ∞ as z → ∞, then f (z) is a polynomial. 4. A nonconstant polynomial always has a root, per the fundamental theorem of algebra. 5. Remember the different types of singularities and why they’re useful, namely poles of order k and essential singularities. ˆ In other words, they have a 6. Meromorphic functions are holomorphic functions on C. finite set of removable singularities, and can always be expressed as as a ratio of two holomorphic functions. 7. A weak formulation of Picard’s little theorem tells us that if f is entire, then its image is dense. Evaluating integrals: 1. There are generally two ways we’re learned to do nontrivial integrals: 1. Use the closed curve theorem to give a contour integral as zero, then decompose the contour and bound the individual parts to find a specific integral. 2. Use the residue theorem by first calculating residues: 95
(0) Compute the integral
R C (α)
f (w)dw
(1) Compute the Laurent series for f centered at α, and take c−1 (2) If f has a simple pole, then c−1 = limz→α (z − α)f (z) (3) If f (z) = A(z)/B(z) for B having a simple pole and A 6= 0, then c−1 = Res(f ; α) = A(α)/B 0 (α). The value of the integral over a curve containing these poles is 2πi × (sum of residues). Series: 1. Know the definitions of radius of convergence, lim|an |1/n = L, and convergence tests/arguments for power series. These may be things that range from proving that a power series of a holomorphic function is uniformly convergent to finding the radii of convergence of some functions. 2. Laurent series are power series that extend in both −∞ and ∞ directions, and are typically defined on annuli (why?). To compute Laurent series, you can generally 1. Start with known power series and make the appropriate substitutions/adjustments. 2. Use known geometric series. Being able to decompose a function into its partial fraction expansion is a good trick to know here. Others: 1. Know the topological aspects: polygonally connected (path connected), polygonally simply connected (simply connected), what is meant by a region, etc. Note any open path connected set is simply connected. R R PP PR 2. Know when to interchange , , ∂ , lim , etc. To deal with interchange of sums, show absolute convergence. To deal with interchange of limits, show uniform convergence. (This is covered in real analysis, and also in baby Rudin.) 3. A helpful tool is that you can bound the absolute value of any integral over CR by 2πR times the integrand; this is known as the ML theorem or the estimation lemma.
96
APPENDIX
B ON THE FOURIER TRANSFORM
This is an addendum to the class notes on Fourier transforms, and is based on Chapter 4 of Stein and Shakarchi’s Complex Analysis. It is important to note that a function f : R → C must satisfy appropriate regularity and decay conditions to possess a Fourier transform. If it does (we’ll make this notion precise below), then the Fourier transform is defined by ∞
Z
fˆ(ξ) =
f (x)e−2πixξ dx,
ξ ∈ R.
(B.1)
fˆ(ξ)e2πixξ dξ,
x ∈ R.
(B.2)
−∞
The Fourier inversion formula is then Z
∞
f (x) = −∞
There is a deep and natural connection between complex analysis and the Fourier transform. For example, given a function f initially defined on the real line, the possibility of extending it to a holomorphic function is closely related to the rapid decay at infinity of its Fourier transform fˆ. 2
Proposition. The functions f (x) = e−πx and f (x) = forms. Proof. Show this yourself.
1 cosh πx
are their own Fourier trans
Definition. For α > 0, denote by Fα the class of all functions that satisfy the following two conditions: (i) f is holomorphic in the horizontal strip Sα = {z ∈ C : |Im(z) < a}. (ii) There is a constant A > 0 : |f (x + iy)| ≤
A 1 + |x|1+
for any > 0 and all x ∈ R and |y| < a. This condition is known as moderate decay. As we will soon see, Fα can be regarded as the class of functions with well-defined Fourier 97
transforms. Intuitively, one can think of Fα as those holomorphic functions of Sα that are of “moderate decay” on each horizontal line Im(z) = y, uniformly in −a < y < a. Some functions that belong to Fα include 2 f (z) = e−πz for all α, and f (z) =
1 c 2 π c + z2
for 0 < a < c. Proposition. If f ∈ Fα , then f (n) ∈ Fβ for 0 < β < α. Proof. Use the Cauchy integral formula.
Proposition. If f ∈ Fα for some α > 0, then |fˆ(ξ)| ≤ Be−2πβ|ξ| for some B and any 0 ≤ β < α. Proof. Use contour integration.
Let’s drop the α in Fα to make notation simpler. The previous proposition tells us that if f ∈ F, then fˆ has rapid decay at infinity. The Fourier inversion formula then works in the following sense: Proposition. If f ∈ F, then the Fourier inversion formula holds, and we have Z ∞ f (x) = fˆ(ξ)e2πixξ dξ −∞
for all x ∈ R.
Properties of the Fourier transform. Now that we have some basis for knowing when a Fourier transform or inversion is welldefined, let’s talk about some of the things we do with them. Theorem. (Basic properties) Let f, g ∈ F. Then the following are true: ˆ = afˆ + bˆ (1) If h(x) = af (x) + bg(x) for a, b ∈ C, then h g. 0 0 0 ˆ (2) If h(x) = f (x − x ) for x ∈ R, then f (ξ) = e−2πix ξ fˆ(ξ). 0 ˆ (3) If h(x) = e2πixξ f (x) for ξ 0 ∈ R, then h(ξ) = fˆ(ξ −ξ 0 ). 1 ˆ ξ ˆ (4) If h(x) = f (ax) for a ∈ R − {0}, then h(ξ) = f . |a|
a
ˆ (5) If h(x) = f (x), then h(ξ) = fˆ(−ξ). (6) Let F denote the Fourier transform. F 2 (f )(x) = f (−x) and F 4 (f ) = f . Proof. These are all easy to verify.
98
Theorem. (Relation to differentiation) F
d dx
m
n d m ((−ix) f ) = (iξ) fˆ, dξ n
where F denotes the Fourier transform. In particular, the Fourier transform of (−ix)f (x) d ˆ is dξ f (ξ), and the Fourier transform of f 0 is (iξ)fˆ(ξ). Proof. Use differentiation under the integral sign and integration by parts. Theorem. (Relation to convolution) If f ∗ g denotes the convolution of f and g defined by Z ∞ (f ∗ g)(ξ) := f (ξ)g(ξ − t)dt, −∞
then f[ ∗ g(ξ) = fˆ · gˆ. This latter theorem is important for functional analysis, but we won’t discuss it to make this note self-contained. We can also prove an important tool known as the Poisson summation formula: Theorem. (Poisson summation formula) If f ∈ F, then X X f (n) = fˆ(n). n∈Z
(B.3)
n∈Z
Proof. Let f ∈ Fα and choose a β : 0 < β < α. The function
1 e2πiz −1
has simple poles
f (z) e2πiz −1
1 2πi
with residue at the integers. So has simple poles at the integers n, with residues f (n)/2πi. Applying the residue formula to the contour γN below,
we get X |n|≤N
Z f (n) = γN
f (z) dz. −1
e2πiz
Letting N → ∞, and recalling the moderate decay of f , we see that the sum converges to P f (n), and that the integral over the vertical segments goes to 0. So in the limit we n∈Z get Z Z X f (z) f (z) f (n) = dz − dz, (B.4) 2πiz − 1 2πiz − 1 e e L1 L2 n∈Z
99
where L1 and L2 are the real line shifted down and up by b, respectively. Now we use the fact that if |w| > 1, then ∞ X 1 = w−1 w−n w−1 n=0 to see that on L1 , where |e2πiz | > 1, we have 1 e2πiz − 1 If |w| < 1, then
1 w−1
=−
P∞
= e−2πiz
∞ X
e−2πinz .
n=0
n
n=0
w , so on L2 we have 1 e2πiz − 1
=−
∞ X
e2πinz .
n=0
Substituting these equalities into equation (4) above, we see that ! ! Z Z ∞ ∞ X X X −2πiz −2πinz 2πinz f (n) = f (z) e e dz + f (z) e dz n∈Z
=
∞ Z X n=0
L1
f (z)e−2πi(n+1)z dz +
L1
L2
n=0 ∞ Z X n=0
f (z)e2πinz dz =
L2
∞ X
n=0
fˆ(n + 1) +
n=0
∞ X n=0
where we shift L1 and L2 back to the real line by replacing x 7→ x − ib.
fˆ(−n) =
X
fˆ(n),
n∈Z
The Poisson summation formula gives many useful identities in complex analysis and beyond (namely, number theory). For example, the Fourier transform of the function f (x) = 2 2 e−πt(x+a) is fˆ(ξ) = t−1/2 e−πξ /t e2πiaξ . Applying the Poisson summation formula to this pair gives the following identity: ∞ X n=−∞
2
e−πt(n+a) =
∞ X
2
t−1/2 e−πn
/t 2πina
e
.
n=−∞
This identity P∞ can be 2used to prove the following transformation law for the theta function ϑ(t) := n=−∞ e−πn t : ϑ(t) = t−1/2 ϑ(1/t), for t > 0. It can also be used to prove the key functional equation of the Riemann zeta function, which gives its analytic continuation: πs Γ(1 − s)ζ(1 − s) ζ(s) = 2s π s−1 sin 2 Finally, Fourier transforms can tell us much about a function. For example, the following theorem describes the nature of those functions whose Fourier transforms are supported; in particular, it relates the decay properties of a function with the holomorphicity of its Fourier transform: Theorem. (Paley-Wiener) Suppose f is continuous and of moderate decrease on R. Then f has an extension to the complex plane that is entire with |f (z)| ≤ Ae2πM |z| for some A > 0 if and only if fˆ is supported in the interval [−M, M ]. 100
Applications of the Fourier transform. One can think of the Fourier transform as a mapping that converts a function f (t) in the time domain (t has units of seconds) to a function fˆ(ω) in the frequency domain (ω has units of hertz or something). Fourier transforms can be used to solve differential equations (c.f. below), and is clearly widely used in physics, probability, engineering, and many other areas. For example, one usually proves the central limit theorem in probability theory using Fourier transforms, and can prove a mathematical formulation of the Heisenberg uncertainty principle. For all the Fourier analysis you can get, Stein and Shakarchi’s Fourier Analysis is a good standard textbook (and one you can find online...).
Examples involving the Fourier transform. Here is a flavor of some of the problems one can ask about Fourier transforms. 2
Example 1. This example generalizes some of the properties of e−πx related to that it is its own Fourier transform. Suppose f (z) is an entire function that R∞ 2 2 |f (x + iy)| ≤ ce−ax +by for some a, b, c > 0. Let fˆ(ζ) = −∞ f (x)e−2πixζ dx. 0 2 0 2 is an entire function of ζ that satisfies |fˆ(ζ + iη)| ≤ c0 e−a ζ +b η for some a0 , b0 , c0
the fact satisfies Then fˆ > 0.
d with the Proof. Note that fˆ is clearly an entire function since we can just commute dz −2πixζ ˆ integral, and f (x)e is entire. Thus f does not have any poles, and in particular we can change the contour from the real axis to the line x − iy for some y > 0 fixed and −∞ < x < ∞ without affecting the value of the integral. Assume ζ > 0, and observe that Z ∞ Z ∞ Z ∞ fˆ(ζ) = f (x)e−2πixζ dx = f (x − iy)e−2πi(x−iy)ζ dx = f (x − iy)e−2π(xi+y)ζ dx. −∞
−∞
Then note Z Z ∞ −2π(xi+y)ζ f (x − iy)e dx ≤
−∞
∞
−ax2 +by 2 −2π(xi+y)ζ
ce
e
∞
ce
−ax2 −2πxiζ+by 2 −2πζy
−∞
−∞
−∞
Z dx =
2
dx .
2
Since |e−ax −2πxiζ | → 0 as |x| → ∞, we see that fˆ(ζ) = O(eby −2πζy ). Setting y = dζ where 2 2 2 d is a small constant, we have that fˆ(ζ) = O(eb(dζ) −2πζ(dζ) ) = O(e(bd −2πd)ζ ), as desired. Also observe that Z ∞ Z ∞ −2πix(ζ+iη) −2π(xi+y)(ζ+iη) ˆ |f (ζ + iη)| = f (x)e dx = f (x − iy)e dx −∞
Z =
∞
−∞
−∞
Z f (x − iy)e−2π(xζi−xη+ζy+yηi) dx ≤
∞
−∞
2
ce−ax
+by 2 −2π(xζi−xη+ζy+yηi)
dx
2 2 Repeating the above argument, we see that fˆ(ζ) = O(e(by −2πζy)+(−ax +2πxη) ). Now let y = 2 2 2 2 dζ and x = d0 η for small d, d0 in the bound to see that |fˆ(ζ+iη)| = O(e(bd −2πd)ζ +(−ad0 +2πd0 )η ). 0 2 0 2 So |fˆ(ζ + iη)| ≤ c0 e−a ζ +b η for a0 = 2πd − bd2 , b0 = 2πd0 − ad20 (which are > 0 for small d, d0 ), and some suitable scale factor c0 > 0.
101
Example 2. The problem is to solve the differential equation an
dn dn−1 u(t) + an−1 n−1 u(t) + ... + a0 u(t) = f (t), n dt dt
where a0 , a1 , ..., an are complex constants, and f is a given function. Here we suppose that f has bounded support and is smooth. Proof. We do this inRsteps: ∞ (a) Let fˆ(z) = −∞ f (t)e−2πizt dt. Observe that fˆ is an entire function, and using A integration by parts show that |fˆ(x + iy)| ≤ 1+x 2 if |y| ≤ a for any fixed a ≥ 0. n (b) Write P (z) = an (2πiz) + an−1 (2πiz)n−1 + ... + a0 . Find a real number c so that P (z) does not vanish on the line L = {z : z = x + ic, x ∈ R}. (c) Set Z 2πizt e u(t) = fˆ(z)dz. L P (z) Check that
n X
Z
d aj ( )j u(t) = dt j=0
and
Z
e2πizt fˆ(z)dz =
L
e2πizt fˆ(z)dz
L
Z
∞
e2πixt fˆ(x)dx.
−∞
We then conclude by the Fourier inversion theorem that that the solution u depends on the choice of c.
Pn
j=0
d j ) u(t) = f (t). Note aj ( dt
d (a) fˆ is an entire function because we can commute dz with the integral and the integrand is also an entire function. We know that f ∈ C 2 , so now we use integration by parts: 1 let u = f (t), du = f 0 (t)dt, dv = e−2πizt dt, and v = − 2πiz e−2πizt so that Z ∞ Z ∞ 0 f (t) −2πizt ∞ f (t) −2πizt f (t)e−2πizt dt = − e ]t=−∞ + e dt |fˆ(z)| = 2πiz −∞ −∞ 2πiz
Now let u = f 0 (t), du = f 00 (t)dt, dv =
e−2πizt 2πiz dt,
1 −2πizt and v = − (2πiz) so that 2e
Z ∞ f (t) −2πizt ∞ f 0 (t) −2πizt ∞ f 00 (t) −2πizt . |fˆ(z)| = − e ]t=−∞ + − e ] + e dt t=−∞ 2 2 2πiz (2πiz) −∞ (2πiz) Since fˆ is entire, we can (as in the previous exercise) change the contour of integration since fˆ does not have any poles. So we change the contour from t to t − is, as in the argument on page 115 of the text. Rewriting, we have f 0 (t − is) f (t − is) −2πi(x+iy)(t−is) ∞ −2πi(x+iy)(t−is) ∞ ˆ e ]t=−∞ + e ]t=−∞ |f (x + iy)| ≤ − 2πi(x + iy) (2πi(x + iy))2 Z ∞ f 00 (t + is) −2πi(x+iy)(t−is) e dt + . 2 −∞ (2πi(x + iy)) 102
If f 0 and f 00 also decay, we can bound this again: M1 M0 −2πs(x+iy) −2πs(x+iy) ˆ + e |f (x + iy)| ≤ − e 2 2πi(x + iy) (2πi(x + iy)) Z ∞ M2 −2πs(x+iy) + e dt . 2 −∞ (2πi(x + iy)) Note that the first two terms are bounded, and examining the integral we see that we can extract the z12 term. Since y is also bounded, we have that |fˆ(x + iy)| = O( x12 ), and A can therefore conclude that |fˆ(x + iy)| ≤ 1+x 2 for a suitable scale factor A. (b) P is a polynomial of degree n, so it only has finitely many roots and we can choose such a c so that the line L does not pass through a root. So we can choose c1 = 1, and let cn = cn−1 + 1 if P vanishes with the choice of cn−1 . Then c will be the cn for which P does not vanish, and it’ll surely be found after n iterations. We can find some constraints on c if we notice that we want |P (x + ic)| > 0 and can bound |P (x + ic)| as follows: n−1 n−1 X X |P (x + ic)| ≥ |an ||x + ic|n − |aj ||x + ic|j ≥ |an ||x + ic|n − |aj | |x + ic|n−1 j=1
j=1
for |x + ic| > 1 (we can repeat the same procedure with a different bound of |x + ic| instead of |x + ic|n−1 in the rightmost term if |x + ic| < 1, and again solve accordingly). Since we Pn−1 Pn−1 |a | want |an ||x + ic|n − ( j=1 |aj |)|x + ic|n−1 > 0, we see that |x + ic| > j=1 |anj | . We also Pn−1 |a | know that |x + ic| ≥ |c|, so if |x + ic| > 1 we can choose c > max(1, j=1 |anj | ) to satisfy the inequality and we’re done. In particular, one c for which the polynomial does not vanish is Pn−1 |a | c = max(1, j=1 |anj | ) + 1. (c) We differentiate under the integral sign to see that n X j=0
Z = an L
n
aj (
X d j d ) u(t) = aj ( )j dt dt j=0
(2πiz)n
Z L
n
X e2πizt ˆ f (z)dz = aj P (z) j=0
Z ( L
d j e2πizt ˆ ) f (z)dz dt P (z)
Z Z e2πizt ˆ e2πizt ˆ e2πizt ˆ f (z)dz+an−1 f (z)dz+...+a0 (2πiz)0 f (z)dz (2πiz)n−1 P (z) P (z) P (z) L L Z Z P (z)e2πizt ˆ = f (z)dz = e2πizt fˆ(z)dz, P (z) L L
as desired. Furthermore, we claim that the equality Z Z e2πizt fˆ(z)dz = L
∞
e2πixt fˆ(x)dx
−∞
follows by observing that, as in the previous problem, e2πizt fˆ(z) is clearly an entire function d and thus its integral is also entire by commuting dz . So the integral does not have any poles, and in particular we can change the contour of integration from the line L = {z : z = x + ic, x ∈ R} (where c is chosen so that P (z) does not vanish) to the real axis without affecting the value of the integral. 103
Finally, because we’re applying the Fourier inversion to the Fourier transform fˆ of f and both functions satisfy conditions R ∞ in the hypothesis, we get from the Fourier Pn the decay d j ) u(t) = −∞ e2πixt fˆ(x)dx = f (t), and we are done. Thus inversion theorem that j=0 aj ( dt the solution to the differential equation an
dn−1 dn u(t) + a u(t) + ... + a0 u(t) = f (t), n−1 dtn dtn−1
is Z u(t) = L
e2πizt ˆ f (z)dz, P (z)
for P (z) = an (2πiz)n + an−1 (2πiz)n−1 + ... + a0 and c ∈ R so that P (z) does not vanish on the line L = {z : z = x + ic, x ∈ R}.
104
APPENDIX
C REFERENCES
We cite many sources in these notes. The main course textbook is: J. Bak and D. J. Newman, Undergraduate Texts in Mathematics: Complex Analysis. Springer, 2010. One that is oftentimes referenced (and is a gem of a textbook), is E. M. Stein and R. Shakarchi, Princeton Lectures in Analysis 2: Complex Analysis. Princeton UP, 2003. And of course, the classic textbook is none other than Ahlfors: L. V. Ahlfors, Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable. McGraw-Hill, 1979.
105
AFTERWORD
You survived! If you enjoyed what you saw in Math 113, you should consider Math 213a (graduate complex analysis) or Math 229x (analytic number theory). Math 113 provides adequate (dare I say ample) background for both in most years. 213a deals with complex analysis, but on a graduate level: you’ll revisit some of the topological basics, and learn some nice and beautiful mathematics like infinite product expansions and more on elliptic and meromorphic functions. Math 229x is number theory using complex analysis: you’ll begin with things like the prime number theorem and move on quickly. These courses can vary a bit, so do take our recommendation with caution. If you aren’t pursuing further work in mathematics, we hope that you enjoyed the course and remember how nice complex analysis is. Thanks for being a great class, in terms of problem sets and everything else. Keep in touch with Andy or any of us CA’s if you’d like, and have a great summer!
Felix and Anirudha
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