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Let $P(z)$ be any polynomial . If $m$ is any positive integer and $w=\exp(2\pi i/m)$ prove that

$$\frac{P(1) + P(w) + P(w^2) + \dots + P(w^{m-1})}{m} = P(0)$$

and give a geometric interpretation. Is this true for any $f(z)$? I have tried some manipulations on the numerator but still no result.

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closed as off-topic by colormegone, Leucippus, user91500, SchrodingersCat, user223391 Dec 23 '15 at 5:40

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  • $\begingroup$ Prove your result for a single power $z^k$, you can use the sum of a finite geometric series. $\endgroup$ – mlu Dec 22 '15 at 21:25
  • $\begingroup$ Try with m=1 and P(z)=z. $\endgroup$ – mercio Dec 22 '15 at 21:48
  • $\begingroup$ The edit above by root is incorrect on the last polynomial w is raised to m-1....... $\endgroup$ – herashefat Dec 22 '15 at 22:07
  • $\begingroup$ You need to assume the degree of your polynomial $P$ is smaller than $m$. Otherwise, your statement can fail. E.g. consider the case $P(x) = x^m$. $\endgroup$ – achille hui Dec 23 '15 at 2:30
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A hint:

$$\sum_{k=0}^{m-1} e^{\frac{2 \pi i k}{m}}=0$$

since the summands are $m^{th}$ root of unity.

Write $$P(z) = a_0 +a_1 z+\cdots +a_n z^n,$$ and see what happens when you evaluate $P$ at a power of $w$.

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