Prove that $\ \frac{z_1^k + z_2^k + \cdots + z_n^k}{n} = \left(\frac{z_1 + z_2 + \cdots + z_n}{n}\right)^k$ 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 $z^k$
 A: 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}
$
