If $z_{1},z_{2},z_{3},…,z_{n}$ be the n-vertices of a regular n-sided polygon on a complex plane, then prove that
$$\ \frac{z_1^k + z_2^k + \cdots + z_n^k}{n} = \left(\frac{z_1 + z_2 + \cdots + z_n}{n}\right)^k$$ where $k\in N, k<n$.
Further, if centre of the polygon is $z_0$ then prove that $$\ \sum_{r=1}^nz^k_r=nz^k_0$$
My Attempt
$z_2-z_0=(z_1-z_0)e^{\frac{i2\pi}{n}}$
$z_3-z_0=(z_1-z_0)e^{\frac{i4\pi}{n}}$
$z_4-z_0=(z_1-z_0)e^{\frac{i6\pi}{n}}$
…
But what to do about things like $z^k$
Best Answer
This is just a standard exercise in knowing how to specify the vertices of a general regular $n$-gon, manipulating sums, and using $\sum_{k=0}^{n-1}e^{2\pi i k/n} =\begin{cases} n & k=0\\ 0 & 1 \le k < n\\ \end{cases} $.
Here are all the details.
If the center is at $z_0$ then the vertices are $z_k =z_0+re^{2\pi i (k/n+c)} =z_0+re^{2\pi i c}e^{2\pi i k/n} $ for some $r$ and $c$ so
$\begin{array}\\ \sum_{k=1}^n z_k &=\sum_{k=1}^n (z_0+re^{2\pi i c}e^{2\pi i k/n})\\ &=nz_0+re^{2\pi i c}\sum_{k=1}^n e^{2\pi i k/n}\\ &=nz_0+re^{2\pi i c}e^{2\pi i/n}\sum_{k=0}^{n-1} e^{2\pi i k/n}\\ &=nz_0+re^{2\pi i c}e^{2\pi i/n}\dfrac{1-e^{2\pi i n/n}}{1-e^{2\pi i/n}}\\ &=nz_0 \qquad\text{if } n > 1\\ \text{so}\\ \dfrac{\sum_{k=1}^n z_k}{n} &=z_0\\ \\ \text{and} &\text{ for }1 \le m < n \text{ (I use } m \text{ instead of }k \text{ here)}\\ \sum_{k=1}^n z_k^m &=\sum_{k=1}^n (z_0+re^{2\pi i c}e^{2\pi i k/n})^m\\ &=\sum_{k=1}^n (z_0+se^{2\pi i k/n})^m \qquad s = re^{2\pi i c}\\ &=\sum_{k=1}^n \sum_{j=0}^m \binom{m}{j}s^je^{2\pi i jk/n}z_0^{m-j}\\ &=\sum_{j=0}^m \binom{m}{j}s^jz_0^{m-j}\sum_{k=1}^n e^{2\pi i jk/n}\\ &=\sum_{j=0}^m \binom{m}{j}s^jz_0^{m-j}e^{2\pi i j/n}\sum_{k=0}^{n-1} e^{2\pi i jk/n}\\ &=\binom{m}{0}s^jz_0^{m}e^{0}\sum_{k=0}^{n-1} e^{2\pi i 0k/n}+\sum_{j=1}^m \binom{m}{j}s^jz_0^{m-j}e^{2\pi i j/n}\dfrac{1-e^{2\pi i nj/n}}{1-e^{2\pi i j/n}}\\ &=nz_0^m+\sum_{j=1}^m \binom{m}{j}s^jz_0^{m-j}e^{2\pi i j/n}\dfrac{1-e^{2\pi i j }}{1-e^{2\pi i j/n}}\\ &=nz_0^m\\ \text{so}\\ \dfrac{\sum_{k=1}^n z_k^m}{n} &=z_0^m\\ \end{array} $