## De Moivre's Theorem and Complex Number Equations

Consider the rule for multiplying complex numbers in polar form

1= r cis θ1 . r 2 cis θ 2

= r 1cis (θ 1 + θ 2)

### Squaring Complex Numbers

Suppose that the two numbers are equal. i.e. z = z 2

Then z2 = r cis θ. r cis θ = r2cis(θ + θ)

(r cis θ2 = r2 cis(2θ)

### Cubing Complex Numbers

Let z1 = r cis θ and z2 = r cis(2θ)

Then z3 = r cisθ. r cis(2θ) = r cis(θ + 2θ)

(r cis θ3 = r3 cis(3θ)

### De Moivre's Theorem

The above two derivations can be extended into a general rule known as De Moivre's Theorem:

 (r cis θ )n = rn cis( nθ )

Example

Find (3 cis )6

(3 cis )6 = 36 cis 6. = 729 cis

Equations involving Complex Numbers

The aim of this section is to solve equation of the type zn = r cis θ

Taking the nth root of both sides z = (r cis θ)1/n

By De Moivre's Theorem

This provides one of the roots but because complex numbers in polar form repeat every 360° or 2π radians there will be other roots as well.

The full solution set of the nth roots of r cis θ is given by:

This will mean that when finding the nth root there will be n solutions.

Example 1

 Find the three cube roots of z = 8 cis (π) Shown on an Argand diagram:

Note that the three roots are equally spaced around a circle of radius 2 (cube root of 8) on the Argand diagram. This is true for all equations of the form zn = r cis θ. Once one root has been found the other roots can be placed, equally spaced, around a circle of radius the n th root of r.

Example 2

 Solve z4 = –16 Writing -16 in polar form z4 = 16 cis (π)      (Note, the modulus is always positive.) z = (16 cis (π))1/4 For the fourth root there will be 4 solutions. Shown on an Argand diagram: