De Moivre's Formula

De Moivre's Theorem states that for any complex number $x$ and integer $n$,

$( \cos x + i \sin x )^n = \cos ( nx) + i \sin (nx)$

This formula is easily proven by induction on $n$, and applying the trigonometric sum and product formulas. We first deal with the non-negative integers. The base case $n=0$ is clearly true. For the induction step, observe that

$\begin{array} { l l } ( \cos x + i \sin x)^{k+1} & = (\cos x + i \sin x )^k \times ( \cos x + i \sin x ) \\ & = \left( \cos (kx) + i \sin (kx) \right) ( \cos x + i \sin x ) \\ & = \cos (kx) \cos x - \sin(kx) \sin x + i[ \sin (kx) \cos x + \cos(kx) \sin x] \\ & = \cos [(k+1)x] + i \sin [(k+1)x]\\ \end{array}$

Note that this version of the proof only holds true for integers $n$. There is a more general version, in which $n$ is allowed to be a complex number. In this case, the left hand side is a multi-valued function, and the right hand side is one of its possible values.

Euler's Formula for complex numbers states that if $z$ is a complex number with absolute value $r_z$ and argument $\theta_z$, then

$z = r_z e^{i \theta_z}.$

The proof of this is best approached using the (Maclaurin) power series expansion, and is left to the interested reader. With this, we have another proof of De Moivre's theorem that directly follows from the muliplication of complex numbers in polar form.

Worked Examples

1. Evaluate $( 1 + \sqrt{3} i )^{2013}$.

First, we find the absolute value and argument of $z = 1 + \sqrt{3} i$.
Absolute value: $r_z = \sqrt{ 1^2 + \sqrt{3} ^2 } = \sqrt{4} = 2$.
Argument: $\theta_z = \arctan \frac{\sqrt{3} } {1} = \frac{\pi}{3}$

Applying DeMoivre's Theorem, we obtain

$\left[ 2 \text{ cis } \frac{ \pi}{3} \right]^{2013} = 2^{2013} \text{ cis } \frac{ 2013 \pi } { 3} = 2^{2013} ( - 1 + 0 i ) = - 2^{2013}.$

2. Show that $\cos (5\theta) = cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta$.

Applying De Moivre's Theorem for $n= 5$, we obtain

$\cos (5 \theta) + i \sin ( 5 \theta) = ( \cos \theta + i \sin \theta) ^ 5 .$

Expand the RHS using the Binomial Theorem and compare real parts to obtain:

$\cos ( 5 \theta) = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.$

Note: For an integer $n$, we can express $\cos ( n \theta)$ solely in terms of $\cos \theta$ by using the identity $\sin^2 \theta = 1 - \cos^2 \theta$. This is known as the Chebyshev polynomial of the first kind.

3. Evaluate $\sin (0\theta) + \sin (1 \theta) + \sin (2 \theta) + \ldots + \sin (n \theta)$.

Applying De Moivre's Formula, this is equivalent to the imaginary part of

$( \cos \theta + i \sin \theta)^0 + ( \cos \theta + i \sin \theta)^1 + ( \cos \theta + i \sin \theta) ^2 + \ldots + ( \cos \theta + i \sin \theta)^n.$

Interpreting this as a Geometric Progression, the sum is

$\frac{ (\cos \theta + i \sin \theta)^{n+1} -1} {( \cos \theta + i \sin \theta) - 1 },$

as long as the ratio is not 1, which means $\theta \neq 2k \pi$. (Note that in this case, we get that each term $\sin (k\theta)$ is 0, and hence the sum is 0.)

Converting this to polar form, we obtain

$\frac{ e^{i (n+1) \theta} - 1 } { e^{i\theta} -1 } = \frac{ e^{ i \left( \frac{n+1}{2} \right)\theta} } {e^{i \frac{1}{2} \theta} } \times \frac{e^{ i \left( \frac{n+1}{2} \right)\theta} - e^{ - i \left( \frac{n+1}{2} \right)\theta} } { e^{ i \frac{1}{2} \theta} - e^{-i \frac{1}{2} \theta} } = e^{ i\frac{n}{2} \theta} \frac{2i \sin [ ( \frac{n+1}{2})\theta ] } { 2i \sin (\frac{1}{2} \theta)} .$

Taking imaginary parts, we obtain

$\frac{ \sin \left( \frac{n}{2} \theta \right) \sin \left( \frac{n+1}{2} \theta \right) } { \sin \left( \frac{1}{2} \theta \right) }.$