De Moivre's Theorem

DeMoivre's Theorem is a very useful theorem in the mathematical fields of complex numbers. It allows complex numbers in polar form to be easily raised to certain powers. It states that for $x\in\mathbb{R}$ and $n\in\mathbb{Z}$ , $\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)$ .

Proof

This is one proof of De Moivre's theorem by induction.

If $n>0$ , for $n=1$ , the case is obviously true.

Assume true for the case $n=k$ . Now, the case of $n=k+1$ :

$\begin{align*} (\cos x+i \sin x)^{k+1} & =(\cos x+i \sin x)^{k}(\cos x+i \sin x) & \text { by Exponential laws } \\ & =[\cos (k x)+i \sin (k x)](\cos x+i \sin x) & \text { by the Assumption in Step II } \\ & =\cos (k x) \cos x-\sin (k x)+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\ & =\operatorname{cis}(k+1) & \text { Various Trigonometric Identities } \end{align}$ (Error compiling LaTeX. Unknown error_msg)

Therefore, the result is true for all positive integers $n$ .

If $n=0$ , the formula holds true because $\cos(0x)+i\sin (0x)=1+i0=1$ . Since $z^0=1$ , the equation holds true.

If $n<0$ , one must consider $n=-m$ when $m$ is a positive integer.

And thus, the formula proves true for all integral values of $n$ . $\Box$

Note that from the functional equation $f(x)^n = f(nx)$ where $f(x) = \cos x + i\sin x$ , we see that $f(x)$ behaves like an exponential function. Indeed, Euler's identity states that $e^{ix} = \cos x+i\sin x$ . This extends De Moivre's theorem to all $n\in \mathbb{R}$ .

Generalization

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De Moivre's Theorem

Proof

Generalization