$n$ complex numbers with modulus $1$

Question

The problem: Let $z_1$,$z_2$,...$z_n$ $(n \geq 3)$ be complex numbers such that $\left| z_1 \right|=\left| z_2 \right|=\ldots=\left| z_n \right|=1$. Then show that the following statements are equivalent:

1. These are the vertices of a regular polygon.
2. ${z_1}^n+{z_2}^n+\ldots+{z_n}^n=n(-1)^{n+1} z_1 z_2 \ldots z_n$

My approach: We can visualize the numbers as points on a circle with radius $1$. Now, if $(1)$ is true, then difference of arguments between two adjacent vertices would be equal to $\frac{2 \pi}{n}$. W.L.O.G., we can assume that $\arg(z_1)<\arg(z_2)<\ldots <\arg(z_n)$. Let $z_1=e^{i\theta}$. Then, $z_2=e^{i(\theta+\frac{2\pi}{n})}=e^{i \theta}\cdot e^{\frac{2\pi i}{n}}$ and $z_r=e^{i \theta}\cdot e^{\frac{2\pi i(r-1)}{n}}\; \text{for}\; r=1,2,\ldots n$.

So, $$\begin{align} {z_1}^n+{z_2}^n+\ldots+{z_n}^n &= e^{in\theta}(1+1+\ldots+1)\\ &=ne^{in\theta}\end{align}$$

And, $$\begin{align} n(-1)^{n+1} z_1 z_2 \ldots z_n &= n(-1)^{n+1}e^{in\theta}\cdot e^{\frac{n(n+1)}{2}\cdot \frac{2\pi i}{n}}\\ &= n(-1)^{n+1}e^{in\theta}\cdot {(e^{i\pi})}^{(n+1)}\\ &= n(-1)^{n+1}e^{in\theta}\cdot (-1)^{(n+1)}\\ &= ne^{in\theta}\cdot \end{align}$$ So, I've managed to prove that the first condition implies the second. But, I've no idea how to prove it the other way around. Any help would be qppreciated.

Answer 1

Let us assume $z_1^n + .. + z_n^n = n(-1)^{n+1}z_1..z_n $. Then, $|z_1^n + .. + z_n^n| = n = |z_1^n| + .. + |z_n^n| $.
So, since we are summing $n$ complex numbers of module $1$, then $ z_1^n, .., z_n^n $ all have the same argument.

I let you finish...