Q.Prove by method of induction, for all $n \in N$: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$.

Let $P(n):(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. Base: $n=1$ is trivially true. Hypothesis: assume $(\cos\theta+i\sin\theta)^k=\cos k\theta+i\sin k\theta$. Step: $(\cos\theta+i\sin\theta)^{k+1}=(\cos\theta+i\sin\theta)^k(\cos\theta+i\sin\theta)=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)$ (by the hypothesis) $=(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)=\cos(k+1)\theta+i\sin(k+1)\theta$, using the st …

Multiply the assumed form by $(\cos\theta+i\sin\theta)$, expand using $i^2=-1$, and r …

Sign errors from $i^2=-1$ when collecting real and imaginary parts; forgetting to inv …