# De Moivre's Formula Thanks to Abraham de Moivre we have this useful formula:

[ r(cos θ + i sin θ) ]n = rn(cos nθ + i sin nθ)

But what does it do?

It lets us multiply a complex number by itself (as many times as we want) in one go!

Here are the details:

## Complex Numbers

Firstly, a Complex Number is a combination of a Real Number and Imaginary Number:

A Real Number is the type of number we use every day.

Examples: 12.38, ½, 0, −2000

An Imaginary Number, when squared gives a negative result: The "unit" imaginary number when squared equals −1

i2 = −1

### Examples of Complex Numbers:

 3.6 + 4i (real part is 3.6, imaginary part is 4i) −0.02 + 1.2i (real part is −0.02, imaginary part is 1.2i) 25 − 0.3i (real part is 25, imaginary part is −0.3i)

## In Polar Form

A complex number
(example: 3 + 4i) ... can also be in in polar form
(distance and angle) In other words the complex number 3 + 4i can also be shown as distance 5 and angle 0.927 radians.

How do we do the conversions?

From Cartesian to Polar:

• r = √(x2 + y2) = √(32 + 42) = √25 = 5
• θ = tan-1 (y/x) = tan-1 (4/3) = 0.927 (to 3 decimals)

And back again:

• x = r cos( θ ) = 5 × cos( 0.927 ) = 5 × 0.6002... = 3 (at perfect accuracy)
• y = r sin( θ ) = 5 × sin( 0.927 ) = 5 × 0.7998... = 4 (at perfect accuracy)

In fact, a common way to write a complex number in Polar form is

x + iy = r(cos θ + i sin θ)

And "cos θ + i sin θ" is often shortened to "cis θ", so:

x + iy = r cis θ

cis θ is just shorthand for cos θ + i sin θ

So we can write:

3 + 4i = 5 cis 0.927

## De Moivre

The de Moivre formula (without a radius) is:

(cos θ + i sin θ)n = cos nθ + i sin nθ

And including a radius r we get:

[ r(cos θ + i sin θ) ]n = rn(cos nθ + i sin nθ)

The key points are that:

• the magnitude becomes rn
• the angle becomes
And it looks super neat in "cis" notation:

(r cis θ)n = rn cis nθ

Let us use it!

### Example: What is (1+i)6 ?

First convert 1+i to Polar:

• r = √(12 + 12) = √2
• θ = tan-1 (1/1) = π4

In "cis" notation it is now: √2 cis π4 Use the de Moivre formula with an exponent of 6:

(√2 cis π4)6 = (√2)6 cis 6π4

Which simplifies to:

8 cis 3π2

In other words: the magnitude is now 8, and the angle is 3π2 (=270°)

Which is also 0−8i (see diagram)

Note: using Algebra we can come up with the same answer:

• Firstly: (1 + i)2 = 1 + 2i + i2 = 1 + 2i − 1 = 2i
• Then: (1 + i)6 = (2i)3 = 8i3 = −8i

We can prove de Moivre’s Formula using Mathematical Induction:

### HARD Example: Proving De Moivre’s Formula

1. Show it is true for n=1

[r(cos θ + i sin θ) ]1 = r1(cos θ + i sin θ) is True

2. Assume it is true for n=k

[ r(cos θ + i sin θ) ]k = rk(cos kθ + i sin kθ) is True (our assumption!)

Now, prove it is true for "k+1"

In other words we want to prove:

[ r(cos θ + i sin θ) ]k+1 = rk+1[ cos(k+1)θ + i sin(k+1)θ ]

The left hand side can be factored into:

[ r(cos θ + i sin θ) ]k  r(cos θ + i sin θ)

We can use the assumption above that [ r(cos θ + i sin θ) ]k = rk(cos kθ + i sin kθ), so we get:

rk(cos kθ + i sin kθ)  r(cos θ + i sin θ)

Bring rk and r together:

rk+1 (cos kθ + i sin kθ) (cos θ + i sin θ)

Now let us ignore rk+1  for the moment:

(cos kθ + i sin kθ) (cos θ + i sin θ)

cos kθ(cos θ + i sin θ) + i sin kθ(cos θ + i sin θ)

cos kθ cos θ + i cos kθ sin θ + isin kθ cos θ + i2 sin kθ sin θ

cos kθ cos θ + i cos kθ sin θ + isin kθ cos θ − sin kθ sin θ

cos kθ cos θ − sin kθ sin θ + i(cos kθ sin θ + sin kθ cos θ)

Now use the Trigonometric Identities:

sin(A+B) = sin(A)cos(B) + cos(A)sin(B)

cos(A+B) = cos(A)cos(B) − sin(A)sin(B)

With A = kθ and B = θ we get:

cos(kθ+θ) + i sin(kθ+θ)

cos((k+1)θ) + i sin((k+1)θ)

Now bring back the rk+1:

rk+1 [ cos((k+1)θ) + i sin((k+1)θ ]

The left hand side has been turned into the right hand side!

And so:

[ r(cos θ + i sin θ) ]k+1 = rk+1[ cos(k+1)θ + i sin(k+1)θ ]

is True

DONE!

## Euler's Formula We can create de Moivre's Formula with some help from Leonhard Euler!

eix = cos x + i sin x

And the exponent laws let us do this:

(e)n = einθ

Now replace e with cos θ + i sin θ, and einθ with cos nθ + i sin nθ:

(cos θ + i sin θ)n = cos nθ + i sin nθ

And including a radius r:

[ r(cos θ + i sin θ) ]n = rn(cos nθ + i sin nθ)

Ta Da! We get de Moivre's formula.