Let $n\in\mathbb{N}$, $z\in\mathbb{C}$, and $\zeta_k=\exp\left(\frac{2\pi i k} {n}\right)$. What is $\sum_{k=0}^{n-1}|z-\zeta_k|^2?$
I found this problem in a homework about complex numbers and I couldn't see how to simplify the result. I know that $|z-\zeta_k|^2=|z|^2+|\zeta_k|^2-2\operatorname{Re}(z\bar{\zeta_k})$, and I also know that $|\zeta_k|=1$. Yet somehow I can't figure how to simplify it. Every time I calculate it I get a wrong answer.
Using what you know already
$$ \begin{align} \sum_{k=0}^{n-1}|z-\zeta_k|^2&=\sum^{n-1}_{j=0}(z-\zeta_k)(\overline{z}-\overline{\zeta_k})=\sum_{k=0}^{n-1} \big(|z|^2-2\operatorname{Re}(\overline{z}\zeta_k)+|\zeta_k|^2\big)\\ &=n(|z|^2+1) -2\sum^{n-1}_{k=1}\operatorname{Re}(\overline{z}\zeta_k) \end{align} $$
Using the properties of the real part function and the fact that $\sum^{n-1}_{k=0}\zeta_k$ is a geometric we get
$$ \sum^{n-1}_{k=1}\operatorname{Re}(\overline{z}\zeta_k)=\operatorname{Re}\left(\overline{z}\sum^{n-1}_{k=0}\zeta_k\right)=\operatorname{Re}\left(\overline{z}\sum^{n-1}_{k=0}e^{\tfrac{i2\pi}{n}k}\right)=\operatorname{Re}\left(\overline{z}\frac{1-e^{i2\pi}}{1-e^{\tfrac{i2\pi}{n}}}\right)=0 $$
Putting things together
$$ \sum_{k=0}^{n-1}|z-\zeta_k|^2 = n(|z|^2+1) $$