# Proof of complex numbers

Let $w \in \mathbb C$ where $|w|=1$.

I am trying to prove that there exists $\theta \in \mathbb R$ such that $- \frac{i}{2}(w^n-w^{-n})=\sin(n\theta)$ for all $n\in \mathbb N$

To begin, I thought about using DMT which states that

$(\cos\theta+i\sin\theta)^n= \cos(n\theta)+i\sin(n\theta),$ for any $\theta\in\mathbb R$, $n\in\mathbb Z$ but I am not exactly sure how to use this proposition in the proof.

Any help is appreciated!

If you put $w=\cos\theta + i\sin\theta$, then as you have noted, we have $w^n = \cos(n\theta) + i\sin(n\theta)$, and also $w^{-n} = \cos(n\theta) - i\sin(n\theta)$. Hence $$-\frac{i}{2}(w^n - w^{-n}) = -\frac{i}{2}(2i\sin(n\theta)) = \sin(n\theta),$$ so it turns out that you can pick $\theta$ to be the argument of the complex number $w$.
Hint: Multiplication by $w$ is a rotation of a certain angle $θ$ about the origin because $|w| = 1$. So multiplication by $w^n$ is of course a rotation of $nθ$ about the origin. Also, $w^{-1}$ is the reflection of $w$ about the real axis because $|w| = 1$. Same for $w^{-n}$ and $w^n$.
Given that: $- \frac{i}{2}(w^n-w^{-n})=\sin(n\theta)$ for all $n\in \mathbb N$
So we can say that $$-2i\cdot \frac{i}{2}(w^n-w^{-n})=2i \sin(n\theta)$$ $$(w^n-w^{-n})=\left(\cos n\theta + i\sin n\theta \right) - \left(\cos n\theta - i\sin n\theta \right)$$ $$(w^n-w^{-n})=\left(\cos \theta + i\sin \theta \right)^n - \left(\cos \theta + i\sin \theta \right)^{-n}$$
So it is evident from above that: $$w=\cos \theta + i\sin \theta$$ and hence $\theta$ is the argument of the complex number $w$.