$w^k$ and $w^{n-k}$ are conjugate and inverse of each other

Question

This is my first proof class and I'm struggling with the complex chapter. Please tell me if my proof is correct, or needs help.

Let $w=e^{2\pi i/n}$ where $n\in \mathbb{Z}>0$ and prove that $w^k$ and $w^{n-k}$ are conjugate and inverse of each other $\forall k\in\mathbb{Z}:0\leq k\leq n-1$.

I found that,

$w^k=e^{2k\pi i/n}\implies \cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}$.

Also,

$w^{n-k}=e^{\frac{2\pi i}{n}\cdot n-k}=e^{2\pi i}e^{-\frac{2k\pi i}{n}}=e^{-\frac{2k\pi i}{n}}=\cos\left(-\frac{2k\pi}{n}\right)+i\sin\left(-\frac{2k\pi}{n}\right)=\cos\left(\frac{2k\pi}{n}\right)-i\sin\left(\frac{2k\pi}{n}\right)$ .

So, $w^k$ and $w^{n-k}$ are conjugates. However, this is where I am a little confused, to prove they are inverse of each other can I simply say:

\begin{equation} \frac{1}{w^k}=w^{n-k}\implies \frac{1}{w^k}=w^nw^{-k}\implies w^n=1 \end{equation}

which is true; therefore, the equality $\frac{1}{w^k}=w^{n-k}$ is also true by direct proof?

Answer 1

The first part is perfect.

For the second one, $w^k w^{n-k}= e^{\frac{2\pi k i}{n}}e^{\frac{2\pi (n-k)i}{n}}=e^{\frac{2\pi k i}{n} +\frac{2\pi (n-k)i}{n}}=e^{\frac{2\pi n i}{n}}=e^{2\pi i}=1$

Answer 2

Hint: If $a$ and $b$ are multiplicative inverses of each other then $a.b=1=$(multiplicative identity).