# De Moivre’s Theorem

De Moivre’s theorem is used in the computation of the $n^{\text{th}}$ power of a complex number. In this theorem, the polar (or trigonometric) form of the complex number is used.

De Moivre’s theorem states that for any positive integer $n,$ $[r(\cos\theta+i\sin\theta) ]^n=r^n(\cos n\theta+i\sin n \theta)$ $\text{or,}\;\;\; (\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta$

Proof: We prove this theorem by using mathematical induction.

If $n=1$, then, $[r(\cos\theta+i\sin\theta)]^1 = r(\cos\theta+i\sin\theta)$

Therefore, theorem is true for $n=1$.

If $n=2$, then, $\begin{array}{l} & [r(\cos\theta+i\sin\theta)]^2 \\ =& r^2(\cos^2\theta+i^2\sin^2\theta+2i\sin \theta\cos\theta) \\ =& r^2 (\cos^2\theta-\sin^2\theta+i\sin 2\theta) \\ =& r^2(\cos 2\theta+i\sin 2\theta)\end{array}$

Therefore, theorem is true for $n=2$.

Suppose, this theorem is true for $n=k$ $[r(\cos\theta+i\sin\theta)]^k=r^k(\cos k\theta+i\sin k\theta)$

Multiplying both sides by $r(\cos\theta+i\sin\theta)$, we get, $\begin{array}{l} [r(\cos\theta+i\sin\theta) ]^{k+1} \\ \;\;\;\;\; = r^{k+1}(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta) \\ \;\;\;\;\; =r^{k+1}[\cos(k\theta+\theta)+i\sin(k\theta+\theta)] \\ \;\;\;\;\; =r^{k+1}[\cos(k+1)\theta+\sin(k+1)\theta] \end{array}$

Thus, the theorem is true for $n=k+1$ whenever it is true for $n=k$. Hence, by mathematical induction, the theorem is true for all positive integers $n$. In fact, the theorem is true not only for a positive integer but also for any real number $n$.

De Moivre’s theorem is applied to compute the integral powers of complex numbers.