If $z_1,z_2,z_3, \ldots ,z_n$ are the vertices of an $n$-sided regular polygon with $z_0$ as its centre, then find $$\sum_{r=1}^{n} z_r^k$$where $k \in \Bbb N, k<n$.
And my working is here:
$$z_r - z_0 = (z_1-z_0)e^{i(2\pi(r-1)/n)}$$ $$\sum_{r=1}^n z_r^k = \left(z_0+(z_1-z_0)e^{i(2\pi(r-1)/n)}\right)^k$$
Thanking you in anticipation, Ann
As recommended in the comments start with the Lemma:
$$\forall\{m:\, n\nmid m\}:\quad \sum_{r=0}^{n-1}\left(\omega_{n,\alpha}^r\right)^m=0,\tag1$$ where $\omega_{n,\alpha}^r=e^{i\left(\alpha+\frac{2\pi}nr\right)}$
The proof is trivial: $$ \sum_{r=0}^{n-1}e^{i\left(\alpha+\frac{2\pi}nr\right)m}=e^{i\alpha m}\sum_{r=0}^{n-1}e^{i\frac{2\pi}nrm}=e^{i\alpha m}\frac{1-e^{i(2\pi m)}}{1-e^{i\frac{2\pi}nm}}=0. $$
Now since $z_r=z+R\omega_{n,\alpha}^r$ we have $$ \begin{aligned} \sum_{r=0}^{n-1}z_r^k&=\sum_{r=0}^{n-1}\left(z+R\omega_{n,\alpha}^r\right)^k\\ &=\sum_{r=0}^{n-1}\sum_{m=0}^k \binom km z^{k-m}R^{m}\left(\omega_{n,\alpha}^r\right)^m\\ &=\sum_{m=0}^k \binom km z^{k-m}R^{m}\sum_{r=0}^{n-1}\left(\omega_{n,\alpha}^r\right)^m\\ &=nz^k. \end{aligned} $$
The last equality holds since in view of $k<n$ only the term with $m=0$ survives according to Lemma $(1)$.