Math 220A Fall 2007 Homework #5 Will Garner A04528276

Will Garner A04528276 Pg 54 #1: Find the image of {z: Re z < 0, |Im z| < p} under the exponential function. ez = eu(x, y) + iv(x, y) = eueiv. The real part of z must be less than zero, hence we have that -¶ < u < 0. Since |Im z| < p, we have that -p < v < p, which completes a rotation about a circle. Hence, we have that the image will be all circles of radii eu for -¶ < u < 0. More precisely, it is the punctured, open unit disk. ■ Visually, we have the following:

p

1

-p

Figure 1: Visual depiction of the image of eS if S = {z: Re z < 0, |Im z| < p}

#7: If Tz =

az + b , find z2, z3, z4 (in terms of a, b, c, d) such that Tz = (z, z2, z3, z4). cz + d

In the cross ratio, we map z2 ö 1, z3 ö 0, and z4 ö ¶. So, we have that:

Tz2 =

az2 + b d −b = 1 ⇒ az2 + b = cz2 + d ⇒ z2 = cz2 + d a−c

Tz3 =

az3 + b b = 0 ⇒ az3 + b = 0 ⇒ z3 = − cz3 + d a

Tz4 =

az4 + b d = ∞ ⇒ cz4 + d = 0 ⇒ z4 = − . ■ cz4 + d c

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Will Garner A04528276 #10: Let D = {z: |z| < 1} and find all Möbius transformations T such that T(D) = D. In Exercise 9, we found that the set of maps which takes the unit circle G to itself is (1 + i )(eiθ2 − eiθ3 )eiθ4 ( z − 1) − 2eiθ3 (eiθ2 − eiθ4 )( z − i ) . given by Tz = (−1 − i )eiθ3 ( z − 1) + 2eiθ4 ( z − i ) − (1 − i )eiθ2 ( z + 1) The Symmetry Principle tells us under this map, then all of the interior points of D will be mapped to either the interior of D or the exterior of D. And by the Orientation Principle, so long as (eiθ2 , eiθ3 , eiθ4 ) follows the same orientation as (-1, 1, i), then the interior of D will be mapped to the interior of D. So, the set of Möbius transformations is given by: ⎧ (1 + i )(eiθ2 − eiθ3 )eiθ 4 ( z − 1) − 2eiθ3 (eiθ2 − eiθ4 )( z − i ) ⎫ Tz :⎪ = ⎪ (−1 − i )eiθ3 ( z − 1) + 2eiθ4 ( z − i ) − (1 − i )eiθ2 ( z + 1) ⎬ . ■ ⎨ ⎪(eiθ2 , eiθ3 , eiθ4 ) has the same orientation as (−1, 1, i ) ⎪ ⎩ ⎭

#14: Suppose that one circle is contained inside another and that they are tangent at the point a. Let G be the region between the two circles and map G conformally onto the open unit disk. (Hint: first try (z – a)-1.)

a

WLOG, suppose the region we are starting with appears to the right. (Otherwise, we could just apply a rotation.) We want to map the region to a horizontal strip. Then, by applying a dilation and a translation, we can map the strip to the strip bounded by 0 < Im(z) < p. (See Figure 3.) Exponentiating this region, we will get the upper half plane. (See Figure 4.) From there, we can apply a Möbius transformation which maps -1 ö -1, 0 ö -i, 1 ö 1. By the orientation principle, this will map the upper half plane to the open unit disk. (See Figure 5.)

Figure 2 p

Figure 3

Figure 4

The question remains: how do we map the region to a horizontal strip? But that can be accomplished with the map (z – a)-1. It sends a to 0 and then maps it to ¶, sending each circle to a horizontal line. And since the circles meet at an angle of 0 degrees, we have that the circles will be parallel, hence creating the necessary horizontal strip. ■

1

Figure 5

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Will Garner A04528276 #24: Let T be a Möbius transformation, T ∫ the identity. Show that a Möbius transformation S commutes with T if S and T have the same fixed points. (Hint: Use Exercises 21 and 22.) By Exercises 21 and 22, we have that if T has fixed points z1 and z2, then S-1TS has fixed points S-1z1 and S-1z2. Also, we have that if 0 and ¶ are the only fixed points, then we have a dilation. If ¶ is the only fixed point, then we have a translation. S and T can only share one or two fixed points. (If they shared more, then the maps would be uniquely defined and would be the identity, which is not permitted.) Suppose T and S have one fixed point. That is, Tz1 = z1 = Tz2. Let W be a Möbius transformation such that W(¶) = z1. (So, W-1(z1) = ¶.) Then, by Exercise 21, we have that W-1TW has ¶ as a fixed point. Similarly, W-1SW has ¶ as a fixed point. By Exercise 22, they are translations, which clearly commute. Hence, W-1TWW-1SW = W-1SWW-1TW, i.e. S(T(z)) = T(S(z)). Similarly, if T and S have two fixed points, let W2 be a Möbius transformation such that W(¶) = z1 and W(0) = z2. (So, W-1(z1) = ¶ and W-1(z2) = 0.) Then, by Exercise 21, W-1TW has ¶ and 0 as fixed points. Similarly, W-1SW has ¶ and 0 as fixed points. By Exercise 22, they are dilations, which clearly commute. Hence, W-1TWW-1SW = W-1SWW-1TW, i.e. S(T(z)) = T(S(z)). ■ Pg 67 #6: Show that if g: [a, b] ö C is a Lipschitz function, then g is of bounded variation. Suppose g is a Lipschitz function, that is, suppose that |g(x) – g(y)| § L|x – y|. Recall, g is of bounded variation if there is a constant M > 0 such that for any m

partition P = {a = t0 < t1 < … < tm = b} of [a, b], v(γ , P) = ∑ γ (tk ) − γ (tk −1 ) ≤ M . k =1

Since g is Lipschitz, |g(tk) – g(tk – 1)| § L|tk – tk – 1| § L|a – b| < ¶. Hence, we have that m

m

k =1

k =1

v(γ , P) = ∑ γ (tk ) − γ (tk −1 ) ≤ ∑ L a − b = mL a − b . m is finite, L is finite and |a – b| is finite. Thus, letting M = mL|a – b|, we have that g is of bounded variation. ■

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Will Garner A04528276 #7: Show that g: [0, 1] ö C defined by γ (t ) = t + it sin

1 for t ∫ 0 and g(0) = 0, is a path, t

but is not rectifiable. Sketch this path. To show that g is a path, we need to show that it is continuous on [0, 1]. Notice that g(t) is well-defined and continuous on the interval (0, 1]. Thus, we need only verify that lim γ (t ) = 0 . t →0

Fix e > 0. We want to find a d > 0 such that 1 γ (t ) − 0 = t + it sin − 0 − 0i < ε for |t – 0| < d. t

3 4 1 2 1 4

1 4

1 2

3 4

1

Figure 6: Path of g(t) on [0, 1]

1 1 t + it sin − 0 − 0i = t 2 + t 2 sin 2 ≤ t 2 + t 2 = 2 t . Thus, we if let d = ε / 2 , then t t 1 we have t + it sin − 0 − 0i < ε . Hence lim γ (t ) = 0 and g(t) is continuous on [0, 1]. t →0 t

To show that g(t) is not rectifiable, we need to find a partition P such that v(g, P) > M for all M. 2 Consider the following partition: t0 = 0, tm = 1 and tm − k = , for k = 1, …, m – 1. π (2k − 1) m

v(γ , P) = ∑ γ (tk ) − γ (tk −1 ) = k =1

2 2 (2(m − 1) − 1)π +i sin − 0 + ... (2(m − 1) − 1)π (2(m − 1) − 1)π 2 +

2

π

+i

2

π

sin

π 2

−

π 2 2 3π 2 2 − i sin + 1 + i sin(1) − − i sin π π 3π 3π 2 2

⎛ 4 5 4 17 4 35 4 4k 2 + 1 = 0.417 + ⎜ + + + ... + ⎜ 3π 15π 35π (2k + 1)(2k − 1)π ⎝ ≥

4 5 4 17 4 35 4 4k 2 + 1 + + + ... + 3π 15π 35π (2k + 1)(2k − 1)π

4 4k 2 + 1 m −1 4(2k ) ≥∑ ≥∑ 2 2 k =1 (4k − 1)π k =1 4k π m −1

=

2

m −1

⎞ ⎟ + ... (written in reverse order) ⎟ ⎠

(Here, we neglect the first and last term)

1 4k 2 + 1 2 k (Since ≥ 2 ⇔ 3 ≥ 2 which is clearly true) 2 4k 4k − 1 4k

1

∑k π k =1

As m ö ¶, we have that the summation diverges. Hence, we have found a partition P such that v(g, P) > M for all M. A sketch of the path appears in Figure 6. ■

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Will Garner A04528276 #9: Define g: [0, 2p] ö C by g(t) = eint, where n is some integer (positive, negative, or zero). Show that

1

∫γ z dz = 2π in .

Since g is rectifiable and differentiable, then by Definition 1.12, we have that the path integral given by the path z = eint can be rewritten as follows:

∫γ

1 dz = z

2π

2π

0

0

− int int ∫ e ( ine ) dt =

∫ ( in ) dt = 2π in

(Note: If z = eint, then dz = ineintdt.) We did not make any use of the fact that n was positive, negative or zero in our derivation, hence the formula holds for all integers n. ■

#12: Let I (r ) = ∫

γ

I (r ) = ∫

π

γ

π

eiz dz , where g: [0, p] ö C by g(t) = reit. Show that lim I (r ) = 0 . r →∞ z π

it

π

it

it eiz ei ( re ) ei ( re ) dz = ∫ it (reit )′dt = ∫ it ireit dt = ∫ ieire dt z re re 0 0 0

π

π

π

π

π

0

0

ire ire Re( ire ) dt = ∫ e − Im( re ) dt = ∫ e − r sin(t ) dt = ∫ e− r sin(t ) dt ∫ ie dt ≤ ∫ ie dt = ∫ e it

0

π

∫e 0

it

0

− r sin( t )

it

0

δ

dt = ∫ e dt + 0  

− r sin( t )

it

0

π −δ

∫δ

e

− r sin( t )

π

dt +

∫e π δ

− r sin( t )

dt , for some small d > 0.  

−

bounded

bounded

On the interval [d, p – d], we have that –rsin(t) § -rsin(d). (See Figure 7.) Hence, we have the middle integral is bounded by pe-rsin(d).

-rsin Hd L

d

p-d

As r ö ¶, we have that the middle term tends towards 0. Since d was arbitrarily small, let us consider the limit as d ö 0. We have that the first and last integrals are tending towards 0. Hence, we have that |I(r)| ö 0, i.e. lim I (r ) = 0 . ■ r →∞

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Figure 7: Graph of –rsin(t) and –rsin(d) on [0, p]

Will Garner A04528276 Additional Problems to Look At and Know Pg 54 #4: Discuss the mapping properties of zn and z1/n for n ¥ 2. (Hint: use polar coordinates.) If we write z = reiq, then we see that zn maps z to rneinq. That is, it takes the point, multiplies the argument by n and replaces the radius with rn (instead of r).

z1/n maps z to with n r . ■

n

reiθ / n . That is, the argument is divided by n and the radius is replaced

#6: Evaluate the following cross ratios: (a) (7 + i, 1, 0, ¶) (b) (2, 1 – i, 1, 1 + i) (c) (0, 1, i, -1) (d) (i – 1, ¶, 1 + i, 0). 7+i−0 (a) (7 + i, 1, 0, ∞) = 7 + i − ∞ = 7 + i . ■ 1− 0 1− ∞ 2 −1 2 4 2 − (1 + i ) (b) (2, 1 − i, 1, 1 + i ) = = 1− i = = 2 + 2i . ■ (1 − i ) − 1 −i 1 − i (1 − i ) − (1 + i ) −2i

0−i −i −2i 0 − (−1) = 1 = = −1 − i . ■ (c) (0, 1, i, − 1) = 1− i 1− i 1− i 1 − (−1) 2 (i − 1) − (1 + i ) −2 (i − 1) − 0 (d) (i − 1, ∞, 1 + i, 0) = = = 1− i . ■ ∞ − (1 + i ) i −1 ∞−0

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Will Garner A04528276 az + b , find necessary and sufficient conditions that T(G) = G, where G is the cz + d unit circle {z: |z| = 1}.

#9: If Tz =

A map is uniquely defined by determining the image of three points. Let us map three points on the unit circle onto the unit circle. That is, send −1 ⎯⎯ → eiθ2 , 1 ⎯⎯ → eiθ3 , and i ⎯⎯ → eiθ4 . Also, recall that (z, z2, z3, z4) = (Tz, Tz2, Tz3, Tz4). So we have that, ( z , −1, 1, i ) = (Tz , eiθ2 , eiθ3 , eiθ 4 ) . This gives: Tz − eiθ3 iθ3 iθ3 iθ 2 Tz − eiθ4 ⇔ Tz − e = (−1 − i )( z − 1)(e − e ) eiθ2 − eiθ3 Tz − eiθ4 −2( z − i )(eiθ2 − eiθ4 ) eiθ2 − eiθ4 ⇔ −2( z − i )(Tz − eiθ3 )(eiθ2 − eiθ4 ) = (−1 − i )( z − 1)(Tz − eiθ4 )(eiθ2 − eiθ3 ) z −1 z −i = −1 − 1 −1 − i

Using Mathematica to solve for Tz, we have: Tz =

(1 + i )(eiθ2 − eiθ3 )eiθ4 ( z − 1) − 2eiθ3 (eiθ2 − eiθ4 )( z − i ) (−1 − i )eiθ3 ( z − 1) + 2eiθ4 ( z − i ) − (1 − i )eiθ2 ( z + 1)

Notice that in the case where eiθ2 = −1 , eiθ3 = 1 , and eiθ4 = i , we get precisely the identity map: Tz =

(1 + i )(−1 − 1)i ( z − 1) − 2(−1 − i )( z − i ) = z.■ (−1 − i )( z − 1) + 2i ( z − i ) + (1 − i )( z + 1)

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Will Garner A04528276 #17: Let G be a region and suppose that f: G ö C is analytic such that f(G) is a subset of a circle. Show that f is constant. Recall, if F is analytic on a set G and Im(F) = 0, then we have that F is constant. Let f: G ö {|z| = 1}. This can be accomplished by sending 1 ö ¶, -1 ö 0, i ö 1. ⎛ z +1 ⎞ ⎛ i +1 ⎞ ⎛ i −1 ⎞ z +1 . This gives us the map ⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎝ z −1 ⎠ ⎝ i −1 ⎠ ⎝ i +1 ⎠ z −1 ⎛ i −1 ⎞ f ( z) + 1 . Then F is analytic and its image is the real axis, i.e. Let F ( z ) = ⎜ ⎟ ⎝ i + 1 ⎠ f ( z) − 1 Im(F) = 0. So, F(z) is constant. Thus f(z) is constant as well, and so we are done. ■ #21: Let T be a Möbius transformation with fixed points z1 and z2. If S is a Möbius transformation, show that S-1TS has fixed points S-1z1 and S-1z2. Suppose T is a Möbius transformation with fixed points z1 and z2. We want to show that S-1z1 and S-1z2 remain fixed under S-1TS. However, this follows immediately from the fact that S-1TS(S-1z1) = S-1T(z1) = S-1z1, since SS-1(z) = z and T(z1) = z1. Similarly, we have that S-1TS(S-1z2) = S-1T(z2) = S-1z2. ■ #22: (a) Show that a Möbius transformation has 0 and ¶ as its only fixed points iff it is a dilation, but not the identity. Suppose that T is a Möbius transformation with 0 and ¶ as its only fixed points. Recall, if T(z) = z, then cz2 + (d – a)z – b = 0. Since T(0) = 0, we have that b = 0. We cannot have d = 0, otherwise at z = 0, our transformation is undefined. Since T(¶) = ¶, c = 0. Hence, T(z) = (az + b)/(cz + d) = (a/d)z, which is a dilation. Now, suppose that T(z) is a dilation. That is, T(z) = rz, where r ∫ 1. Then the only points that are mapped to themselves are 0 and ¶. ■ (b) Show that Möbius transformation has ¶ as its only fixed points iff it is a translation, but not the identity. Suppose that T is a Möbius transformation with ¶ as its only fixed point. By similar reasoning as in (a), we have that c = 0. Hence, T(z) = (az + b)/d. T(z) = z implies (a/d – 1)z = -b/d, i.e. z = -b/(a – d). But since ¶ maps to ¶, we must have that a = d. Hence, T(z) = z + b/d, which is a translation. Now, suppose that T(z) is a translation. That is, T(z) = z – r, where r ∫ 0. It is evident that the only point that can be mapped to itself is ¶. ■

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Will Garner A04528276 #26: (a) Let GL2(C) = all invertible 2 μ 2 matrices with entries in C and let M be the group of Möbius transformations. Define j: GL2(C) ö M by ϕ

( ) = czaz ++ db . Show a c

b d

that j is a group homomorphism of GL2(C) onto M. Find the kernel of j. The group action under GL2(C) is multiplication and the group action under M is composition. So, to show that j is a group homomorphism we need to show ⎛⎛a b1 ⎞ ⎛ a2 b2 ⎞ ⎞ ⎛ a2 z + b2 ⎞ ⎛ a2 z + b2 ⎞ that ϕ ⎜ ⎜ 1 ⎟ + b1 c1 ⎜ ⎟ + d1 , i.e. ⎟⎜ ⎟ ⎟ = a1 ⎜ ⎝ c2 z + d 2 ⎠ ⎝ c2 z + d 2 ⎠ ⎝ ⎝ c1 d1 ⎠ ⎝ c2 d 2 ⎠ ⎠

⎛ ⎛ a1a2 + b1c2

ϕ ⎜⎜

⎝ ⎝ c1a2 + d1c2

a1b2 + b1d 2 ⎞ ⎞ a1 (a2 z + b2 ) + b1 (c2 z + d 2 ) ⎟⎟ = c1b2 + d1d 2 ⎠ ⎠ c2 z + d 2

c1 (a2 z + b2 ) + d1 (c2 z + d 2 ) c2 z + d 2

Simplifying the right-hand side, we have: ⎛ ⎛ a1a2 + b1c2

ϕ ⎜⎜

⎝ ⎝ c1a2 + d1c2

a1b2 + b1d 2 ⎞ ⎞ (a1a2 + b1c2 ) z + (a1b2 + b1d 2 ) , as desired. ⎟⎟ = c1b2 + d1d 2 ⎠ ⎠ (c1a2 + d1c2 ) z + (c1b2 + d1d 2 )

The kernel is the center of GL2(C), that is all scalar multiplies of the I2. ■ (b) Let SL2(C) be the subgroup of GL2(C) consisting of all matrices of determinant 1. Show that the image of SL2(C) under j is all of M. What part of the kernel of j is in SL2(C)? Since ϕ

(

λa λc

λb λd

) = λλczaz ++λλdb = czaz ++ db = ϕ ( ) , we can multiply by any non-

zero scalar. In particular, let λ =

a c

b d

1 . This is non-zero, since by definition, ad − bc

ad – bc is non-zero. But the matrix

(

λa λc

λb λd

) now has determinant equal to 1.

Hence, the image of SL2(C) under j is all of M. Since the kernel of j is all scalar multiplies of the 2 μ 2 identity matrix, the only part of the kernel in SL2(C) is I2 and –I2, as those are the only matrices with determinant equal to 1. ■

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Will Garner A04528276 Pg 67 #1: Let g: [a, b] ö R be non-decreasing. Show that g is of bounded variation and

V(g) = g(b) – g(a). Recall, g is of bounded variation if there is a constant M > 0 such that for any m

partition P = {a = t0 < t1 < … < tm = b} of [a, b], v(γ , P) = ∑ γ (tk ) − γ (tk −1 ) ≤ M . k =1

Since g: [a, b] ö R, we have that Im(g) = 0. And since g is non-decreasing, we have that g(tk) ¥ g(tk – 1) for all k. Hence, we have that m

m

k =1

k =1

v(γ , P) = ∑ γ (tk ) − γ (tk −1 ) = ∑ γ (tk ) − γ (tk −1 ) = γ (tm ) − γ (t0 ) , where the middle terms drop out because we have a telescoping sum. But recall tm = b and t0 = a, so we have that v(g, P) = g(b) – g(a), independent of the partition P. In particular, we have that

V(g) = sup{v(g, P) : P a partition of [a, b]} = g(b) – g(a). We have that V(g) is bounded by some constant M since g: [a, b] ö R, hence both g(b) and g(a) are finite. (So their difference will necessarily be finite as well.) ■

#5: Let g(t) = exp ((-1 + i)t-1) for 0 < t § 1 and g(0) = 0. Show that g is a rectifiable path and find V(g). Give a rough sketch of the trace of g. To show that g is a rectifiable path, we first need to show that it is a path. That is, we need to show that it is continuous on [0, 1]. Notice that g(t) is well-defined and continuous on the interval (0, 1]. Thus, we need only verify that lim γ (t ) = 0 .

1 4 3 16 1 8

t →0

Fix e > 0. We want to find a d > 0 such that 1 1 γ (t ) − 0 = e −1 t cos + ie−1 t sin − 0 − 0i < ε for |t – 0| < d. t t

1 16

-1

16

1 16

1 8

3 16

Figure 8: Path of g(t) on [0, 1]

1 1 e −1 t cos + ie −1 t sin − 0 − 0i = e−1 t + e−1 t ≤ 2e−1 t < ε if δ > −1/ ln(ε 2 / 2) . Hence t t lim γ (t ) = 0 and g(t) is continuous on [0, 1]. t →0

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Will Garner A04528276 1 1⎞ ⎛ ⎝ ⎠ −1 t 1 1 ⎞ e−1 t e ⎛ = 2 (1 − i ) ⎜ cos + i sin ⎟ = 2 t t t⎠ t ⎝

γ (t ) = e1 t ( −1+i ) = e−1 t ⎜ cos + i sin ⎟ t t ⇒ γ ′(t ) = e1 t ( −1+i )

2

⇒ γ ′(t ) =

1 1 1 1⎞ ⎛ ⎜ cos + sin − i cos + i sin ⎟ t t t t⎠ ⎝

2

e −1 t ⎛ 1 1⎞ ⎛ 1 1⎞ 2e −1 t ⎜ cos + sin ⎟ + ⎜ − cos + sin ⎟ = t2 ⎝ t t⎠ ⎝ t t⎠ t2 1

e −1 t 2 . A sketch of the graph appears in Figure 8. ■ dt = 2 t e 0

Thus, V (γ ) = 2 ∫

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