<s>[INST] How to prove De Moivre's theorem inductively It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.
I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:
$$(\cos\theta + i\sin\theta)^{k+1}= (\cos k\theta + i\sin k \theta)(\cos\theta + i\sin\theta)$$
How can I rearrange the left to look like the right?
 [/INST] $$(\cos \theta + i\sin \theta)^k = \cos k\theta + i\sin k\theta$$
$$\text{So for } (\cos \theta + i\sin \theta)^{k+1} = (\cos \theta + i\sin \theta)^k \cdot (\cos \theta + i\sin \theta) = (\cos k\theta + i\sin k\theta)\cdot (\cos \theta + i\sin \theta)$$
$$=\cos k\theta\cos\theta +i\sin\theta\cos k\theta +i\sin k\theta\cos\theta +i^2\sin{k\theta}\sin\theta$$
$$=(\cos{k\theta}\cos\theta - \sin k\theta\sin\theta)+i(\sin\theta\cos k\theta +\sin k\theta\cos\theta)$$
$$=\cos(k+1)\theta + i\sin(k+1)\theta$$
Hence, this proves that $(\cos\theta +i\sin\theta)^n = \cos n\theta + i\sin n\theta$.
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