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 )


Find (3 cis Y12_De_Moivres_Theorem_and_Complex_Number_Equations_01.gif)6

(3 cis Y12_De_Moivres_Theorem_and_Complex_Number_Equations_01.gif)6 = 36 cis 6.Y12_De_Moivres_Theorem_and_Complex_Number_Equations_01.gif = 729 cis Y12_De_Moivres_Theorem_and_Complex_Number_Equations_02.gif

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: