De Moivre's Theorem

In this chapter we are going to learn about a famous identity based on .

Introduction:

The de Moivre's theorem (also known as de Moivre's identity), is named after its proposer Abraham De Moivre.

It states that for any $x \in \mathbb{R}$ and $n \in \mathbb{N}$ the relation:

$(\cos x + i \sin x)^n = \cos nx + i \sin nx$, holds true.

Proof (Using Euler's Formula):

Euler's formula states that

$e^{ix} = \cos x + i\sin x$ $...eqn(i)$

Applying the above formula to $nx$ instead of $x$, we get:

$e^{inx} = \cos nx + i \sin nx$

$\Rightarrow {\left(e^{ix}\right)}^n = \cos nx + i \sin nx$

Substituting $e^{ix} = \cos x + i\sin x$ from $eqn(i)$ above, we get:

$\Rightarrow {\left(\cos x + i\sin x\right)}^n = \cos nx + i \sin nx$

Proof (Using Induction):

Let the relationship $(\cos x + i \sin x)^n = \cos nx + i \sin nx$ be true for some $n = k$

$\therefore (\cos x + i \sin x)^k = \cos kx + i \sin kx$

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Multiplying both sides by $(\cos x + i \sin x)$ we get:

$(\cos x + i \sin x)^k(\cos x + i \sin x) = (\cos kx + i \sin kx)(\cos x + i \sin x)$

$\Rightarrow (\cos x + i \sin x)^{k + 1} = \cos kx \cos x + i\cos x \sin kx + i\sin x \cos kx + i^2\sin kx \sin x$

$\Rightarrow (\cos x + i \sin x)^{k + 1} = \cos kx \cos x - \sin kx \sin x + i(\cos x \sin kx + \sin x \cos kx)$ $\because i^2 = -1$

$\Rightarrow (\cos x + i \sin x)^{k + 1} = \cos (kx + x) + i\sin(kx + x)$ $\because i^2 = -1$

$\Rightarrow (\cos x + i \sin x)^{k + 1} = \cos (kx + x) + i\sin(kx + x)$

$\Rightarrow (\cos x + i \sin x)^{k + 1} = \cos (k + 1)x + i\sin(k + 1)x$

Therefore, the given relationship is true for $n = k + 1$, if it is true for $n = k$

For the base case, we use $n = 1$, and get the $LHS$ as:

$(\cos x + i \sin x)^1 = \cos x + i \sin x$

and the $RHS$ as:

$(\cos 1x + i \sin 1x) = \cos x + i \sin x$

Therefore, by induction it is proven that $(cos x + i \sin x)^n = \cos nx + i \sin nx$ for all $n \in \mathbb{N}$