Complex Numbers Syllabus Ordinary Level

Higher Level

Slide 1

THE ARGANG DIAGRAM

Slide 2

Slide 3



Complex numbers have both a real and an imaginary part.



They are usually written as a+ib, where a is the Real part and b is the imaginary part.



For the complex number 2-4i, the real part is 2 and the imaginary part is -4. (Remember the sign belongs to the number that comes after it)

Slide 4

Slide 5

Slide 6

Slide 7 

Add 2+2i+(2-i)



Add the real parts 2+2=4



Add the imaginary parts 2+(-1)=2-1=1



2+2i+(2-i)= 4+i

Slide 8

Slide 9 Plotting B+2-i

Plotting D+2-i

1-i+2-i= 1+2-i-i= 3-2i

-4-2i+2-i= -2-3i

Plotting C+2-i

-2+3i+2-i= 2i

Slide 10

Slide 11

Slide 12 

All the points are shifted by the same amount. This is a translation.



Moving the points by the same amount the distance and direction from the origin to 2-i

Slide 13

Slide 14 When we subtract a complex number we need to make sure we subtract both the real and imaginary parts 2+2i-(2-i) Real part 2-2=0 Im part 2i-(-i)=2i+i=3i 2+2i-(2-i)=3i

Slide 1

Slide 2 For complex numbers: 

Remember:

( x  3)(2 x  4)

Breakup the first bracket x(2 x  4)  3(2 x  4)

( 2  4i )(3  2i ) 2(3  2i )  4i (3  2i ) 6  4i  12i  8i 2 6  8i  8i 2

2 x 2  4 x  6 x  12

but i 

2 x 2  10 x  12

so i 2  (

1  1 ) 2  1

6  8i  8( 1) 6  8i  8 14  8i

Slide 3 

First we can look at multiplying by i A = 3+2i Multiplying by i i(3+2i) =3i+2i2 =3i +2(-1) =-2+3i A is rotated by 90 anti

clockwise

Slide 4 Multiplying A by –i -i(3+2i) =-3i-2i2 =-3i-2(-1) =2-3i

Multiplying by i rotates anti clockwise

Multiplying by –i rotates clockwise

Slide 5  

 

Multiplying by i rotates by 90 degrees If I multiply again by i it rotates another 90degrees and so on. multiplying by i² or -1 rotates by 180 degrees Multiplying by i³ or –i rotates the point by 270 degrees anti clockwise and multiplying by i4 or by 1 rotates the point afull 360 degrees

Slide 6 



Remember: the modulus is the distance from the complex number to the origin. We can find the modulus using the formula

z  a 2  b2 

Where a complex number is a+ib Multiply each point by 2 to H, I, J and K

Slide 7

(-2+i)(2-2i) =-2(2-2i)+i(2-2i) =-4+4i+2i-2i2 =-4+2+6i =-2+6i

A is rotated by the same amount as the rotation between the Real axis and B and their moduli are multiplied

Slide 1

COMPLEX NUMBERS MODULUS AND COMPLEX CONJUGATE

Slide 2

Conjugate,

z  a  ib

Four points A, B, C, and D are shown with their Complex conjugates A’, B’, C’ and D’. The complex conjugate is the image of the point under axial symmetry in the real line. To find the complex conjugate of any complex number I change the sign of the Imaginary part. The complex conjugate of 2-4i is 2+4i

Slide 3

The modulus The modulus of A is distance from the origin to the point A. Looking at the Modulus as the hypotenuse of a right angled triangle we can use Pythagoras to the modulus. h²=x²+y² The x is the distance along the real axis. The y is the distance along the imaginary axis h²=4²+3² h²=16+9 h²=25 h= 5

Slide 4

Modulus Formula 

For a complex number z=a+ib, it’s modulus

Z  a 2  b2 Eg. z = 2-4i z 

2 2  ( 4 ) 2



4  16



20



4 5

2 2

Slide 1

Slide 2 

We start with a number like

5  5i 2i

 

We need to get rid of the 2-i beneath the line We can do this by multiplying by the complex conjugate of 2-i; 2+i on both the top and the bottom

Slide 3 (5  5i ) (2  i ) (2  i ) (2  i ) 5(2  i )  5i (2  i )  2( 2  i )  i ( 2  i ) 10  5i  10i  5i 2 4  2i  2i  i 2 10  15i  5(1)  4  (1) 10  15i  5  4 1 5  15i  5 5 15   i 5 5  1  3i 

Slide 4

Slide 1

Complex Root

Slide 2 

Complex roots of a quadratic equation



3z²+4z+3=0 Use the formula



 b  b 2  4ac 2a 

a=3, b=4, c=3

Complex roots

Slide 3 

Fill in to the formula 4

4 2  4(3)(3) 2(3)

 4  16  36 6  4   20 6  4  20  1  6  4  4 5 1  6  4  2 5i  4  2 5i  or  6 6 2 5 2 5   i or    i 3 3 3 3  

Complex roots

Higher Level Course Polar Form When using the polar form we will need to change our calculator into radians. CASIO(Shift, Setup, 4) and change back once you’re finished. Instead of placing a point in space based on its x and y coordinates (Cartesian coordinates). We can also place a point based on its angle from the x-axis and its distance from the origin and write the point in the form , where r is the distance form the origin and the angle made with the positive x-axis. The distance from the origin can be found by getting the modulus of the complex number. The angle made with the positive x-axis is called the argument(arg). We can use trigonometry to find the angle but we need to look at the quadrant our point is in to decide if my angle is positive or negative and if the angle is acute or obtuse. Any point in the first two quadrants or with positive y coordinates will give us a positive angle. While any below the x-axis or in the last two quadrants will be negative. The angles in the 2nd and 3rd quadrants (negative x coordinate) will have an obtuse and angle and the angle we find using trigonometry will need to be subtracted from π.

First Quadrant Point

3+i

We can use trigonometry to find the arg

The Argument is r is the modulus r=6

Second Quadrant Point

-3+i

We can use trigonometry to find the arg

The Argument is r is the modulus r=6

because

Third Quadrant Point

- 3-i

The Argument is But the arg is also negative because it is under the x-axis so r is the modulus r=6

) This can also be written as See page 13 in formula and tables book

Fourth Quadrant Point

3-i

The Argument is But the arg is also negative because it is under the x-axis so r is the modulus r=6

) This can also be written as

We use the polar form because it allows us to multiply, divide and get powers of complex numbers quickly. This is because multiplication of complex involves adding the angles and multiplying the moduli (as seen above). If z1= r1[Cos 1++iSin 1] and z2= r2[Cos 2+iSin 2] z1z2=r1r2[Cos

1+ 2)+iSin

1+ 2)]

and Examples: [2(Cos ++iSin )][3(Cos +iSin )]= (2)(3)[Cos

+

)+iSin

=6[Cos

)+iSin

)]

=3[Cos

)+iSin

)]

+

)]

De Moivre’s Theorem ,nЄℕ If z= r [Cos +iSin ] then zn= rn [Cos

+iSin

]

Example Use de Moivre’s theorem to write the following in the form a+ib

To put into a+ib form put And

to get the Im(z)

Ans =15625+i0 =15625

nto calculator(in rad mode) to get Re(z)

You can also be asked to prove identities using deMoivre and expansions Example Prove Using deMoivre For z= z4=Cos

+iSin

but z4=

Putting the Re(z) parts equal to each other

Since and

We have been able to replace all any SinA with CosA

De Moivre can also be used to find the square root, cube root, etc.

Using the fact

that or . However, if we use the usual polar form we get only 1 possible root and not the n roots we should be able to find. To get over this problem we use the general polar form which allows us to add any number of full rotations by adding any number of 2πn.

The general polar form Solve the following z4=-4-4i Change into the general polar form of ω=-4-4i

Arg

But it is in the third quadrant so the arg will be negative and I need to subtract it form π Arg= The general polar form of ω= z4=ω so z=

When n=0

When n=1

When n=2

When n=3

When n=4

This is what the fifth roots of -4-4i look like when plotted. Note how all the points have the same modulus and are all the same angle from each other. The fifth roots are all each other. You can be asked to sketch this.

Proof of De Moivre’s Theorem by Induction for nЄℕ. Induction involves three steps. 1. Show true for n=1 2. Assume true for n=k 3. Prove true for n=k+1 Show that If z=r [Cos +iSin ] then zn= rn [Cos If n=1 then z1= r1 [Cos

+iSin

]

z=r [Cos +iSin ] Assume true for n=k, the If z= r [Cos +iSin ] then zk= rk [Cos

+iSin

Show true for n=k+1 zk+1= r [Cos +iSin ] rk [Cos

+iSin

]

]

+iSin

]

away from

= r rk [Cos +iSin ] [Cos

+iSin

]

=rk+1 [Cos +iSin ] [Cos

+iSin

]

=rk+1 [Cos

+iCos

=rk+1 [Cos

+Cos

Sin +i2 Sin Sin Sin -Sin Sin

] ]

=rk+1 [Cos

+Cos

Sin ]

=rk+1 [Cos

+Cos

Sin ]

From page 14 table and formula book Cos(A+B)= cosAcosB-sinAsinB Sin(A+B)= sinAcosB+cosAsinB =rk+1 [Cos

+

=rk+1 [Cos

+ ]

]

QED

Finding the complex roots of functions In order to find the complex roots of function it is easier to use the conjugate root theorem. This states that if is a root then is also a root. First lets look at what happens when we add and multiple complex numbers and their conjugates.

P(z) is a complex polynomial of the form P(z)=

+

+

+......+

Where z=a+ib and a, b Є R P(z)=0 when z is a root by the factor theorem. P(z)=0 then

=0

P( )=

+

+

+......+

P( )= because

+

+

+......+

P( )=

+

+

+......+

since each a is a constant

P( )= Because P( )= Therefore, P( )=0 and from the factor theorem

.

This means for all z a factor of a polynomial then it’s conjugate is also a factor.

Example -2,z1,z2 are the roots of the cubic equation z3+8=0

z2  2z  4 z  2 z3  0z 2  0z  8 -( z 3  2 z 2 ) -2z 2  0 z -(-2z 2  4 z ) 4z+8 (4 z  8) 0

Use

b  b 2  4ac 2a

on to get the other two factors z 2  2 z  4

(2)  (2) 2  4(4) 2 2  4  16 2 2  12 2 2  i2 3 2 1  i 3 or 1  i 3