I need to prove that $Z^{n}=r^{n}(\cos n\theta +i\sin n\theta)$ is true by Induction. Can someone confirm whether I have done it right or not?

Question

The equation $$Z^{n}=r^{n}(\cos(n\theta) +i\sin(n\theta))\space\space\space\space\space\space\space\space (1)$$ has to be proved by induction. It is given that $$Z=r(\cos(\theta) +i\sin(\theta)).\space$$ I first proved (1) true for $n=1$. This gives $$Z=r(\cos(\theta) +i\sin(\theta)).\space$$ This is true as it was given. Next I assumed (1) to be true for $n=q.$ This implies $$Z^{q}=r^{q}(\cos(q\theta) +i\sin(q\theta)).$$ Then I proved (1) to be true for $n=q+1$. This implied; $$Z^{q+1}=r^{q+1}(\cos(q\theta+\theta)) +i\sin(q\theta+\theta)),$$ $$\implies Z^{q}\cdot Z=(r(\cos(\theta) +i\sin(\theta)))^{q+1},\space by\space De\space Moivre's\space Theorem,$$ $$\implies Z^{q}\cdot Z=(r(\cos(\theta) +i\sin(\theta)))^{q}\cdot r(\cos(\theta) +i\sin(\theta)), $$ $$\implies Z^{q}=(r(\cos(\theta) +i\sin(\theta)))^{q},$$ $$\implies Z^{q}=r^{q}(\cos(q\theta) +i\sin(q\theta)).$$

Hence (1) has been proven true for $n=q+1$.

Answer 1

Well, no. You cannot write like that.

Assume that it holds for $n=q$, i.e. $$Z^q=r^q(\cos(q\theta)+i\sin(q\theta)).$$ Then, we have $$\begin{align}Z^{q+1}&=Z\cdot Z^q\\&=r(\cos\theta+i\sin \theta)\cdot r^q(\cos(q\theta)+i\sin(q\theta))\\&=r^{q+1}((\cos\theta\cos(q\theta)-\sin\theta\sin(q\theta))+i(\cos\theta\sin(q\theta)+\sin\theta\cos(q\theta)))\\&=r^{q+1}(\cos(\theta+q\theta)+i\sin(\theta+q\theta))\\&=r^{q+1}(\cos((q+1)\theta)+i\sin((q+1)\theta)).\end{align}$$ Hence, it holds for $n=q+1$.

Answer 2

The question asks us to prove the validity of the de Moivre Formula, and the proposed proof uses the de Moivre Formula.

For the induction step you are expected to multiply $\cos(n\theta)+i\sin(n\theta)$ by $\cos\theta+i\sin\theta$ in the natural way, and then quote certain trigonometric addition laws.