📂Complex Anaylsis

Proof of De Moivre's Theorem

Theorem

If $z = r \text{cis} \theta$, then for all natural numbers $n$, $z^n = r^n \text{cis} n\theta$ holds.

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• $\text{cis} \theta: = \cos \theta + i \sin \theta$

Proof

Let’s use mathematical induction.

For $n=1$, it is obvious, and assuming it holds for $n=k$, $$z^{k+1} = z z^k = (r \text{cis} \theta)(r^k \text{cis} k\theta)$$ Meanwhile, since $z_1 z_2 = r_1 r_2 \text{cis} (\theta_1 + \theta_2)$, $$z^{k+1} = r^{k+1} \text{cis} (k+1)\theta$$

Since it holds for $n=k+1$ when $n=k$, the given formula holds for all natural numbers.

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