11.3 Powers of Complex Numbers,DeMoivre's Theorem

Objective To use De Moivre’s theorem to find powers of complex numbers.

(r cisθ)³ = (r cisθ)(r cisθ)(r cisθ)

Remember: multiply absolute values and add polar angles.

= (r² cis 2θ) (r cisθ)

= r³ cis 3θ

(r cis θ)ⁿ = rⁿ cis nθ

Abraham de Moivre(1667 - 1754)

DeMoivre’s Theorem

If is a complex number,

then

z r i cos sin

z r n i nn n cos sin integer. positive a is 1 where n

You can repeat this process raising complex numbers to powers. Abraham DeMoivre did this and proved the following theorem:

z r n i nn n cos sin

This says to raise a complex number to a power, raise the modulus to that power and multiply the argument by that power.

6

54sin

6

54cos24

i

43 i

r 3 1 4 22 2

4

4

6

5sin

6

5cos23

ii

This theorem is used to raise complex numbers to powers. It would be a lot of work to find

iiii 3333 you would need to FOIL and multiply all of these together and simplify powers of i --- UGH!Instead let's convert to polar form

and use DeMoivre's Theorem.

3

1tan 1 but in Quad II

6

5

3

10sin

3

10cos16

i

i

2

3

2

116

8 8 3i

Note: In order to use De Moivre’s theorem, the equation must be in polar form.

Solve the following over the set of complex numbers:

13 z We know that if we cube root both sides we could get 1 but we know that there are 3 roots. So we want the complex cube roots of 1.

Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula:

n

k

ni

n

k

nrz n

k

2sin

2cos

1 , ,2 ,1 ,0 where nk

101 22 r

n

k

ni

n

k

nrz n

k

2sin

2cos

Let's try this on our problem. We want the cube roots of 1.

We want cube root so our n = 3. Can you convert 1 to polar form? (hint: 1 = 1 + 0i)

01

0tan 1

We want cube root so use 3 numbers here

Once we build the formula, we use it first with k = 0 and get one root, then with k = 1 to get the second root and finally with k = 2 for last root.

2,1,0for ,3

2

3

0sin

3

2

3

0cos13

k

ki

kzk

10sin0cos1 iHere's the root we already knew.

3

12

3

0sin

3

12

3

0cos13

1

iz

ii2

3

2

1

3

2sin

3

2cos1

3

22

3

0sin

3

22

3

0cos13

2

iz

ii2

3

2

1

3

4sin

3

4cos1

If you cube any of these numbers

you get 1. (Try it and see!)

3

02

3

0sin

3

02

3

0cos13

0

iz

2,1,0for ,3

2

3

0sin

3

2

3

0cos13

k

ki

kzk

ii2

3

2

1,

2

3

2

1,1 We found the cube roots of 1 were:

Let's plot these on the complex plane about 0.9

Notice each of the complex roots has the

same magnitude (1). Also the

three points are evenly spaced

on a circle. This will always be

true of complex roots.

each line is 1/2 unit

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au

Assignment

P. 410 #1, 5, 7, 17