$(1-w_1)\dots (1-w_{n-1})=n?$

by e.turatti   Last Updated September 13, 2017 14:20 PM

To get some context, I was trying to calculate the residue of the function $f(z)=\frac{1}{z^n+1}$ in the point $z_0=e^{\pi i/n}$, so it will be $\frac{1}{(z_0-z_1)\dots(z_0-z_{n-1}) }$, where $z_k=e^{(2\pi ik+\pi i)/n}$, so if I calculate $(z_0-z_k)=e^{\pi i/n}(1-e^{2\pi ik/n})=z_0(1-w_k)$, and $w_k$ is the $k$th root of $1$. So after this I got $$ \frac{1}{(z_0-z_1)\dots(z_0-z_{n-1}) }=\frac{1}{z_0^{n-1}(1-w_1)\dots (1-w_{n-1})}, $$ I couldn't solve this, so I try to look for this question, and I saw other people just saying that $Res(f,z_0)=\frac{1}{nz_0^{n-1}}$.

Looking geometrically I could see that this product seems to be true, but I can't prove this rigorously.

So is this right and how can I try to prove it, or I made some mistake? And maybe there is a better way to calculate this residue?

Thank in advance.



Answers 1


You can factor the polynomial $ z^n + 1 $ in two ways, one being $$ z^n + 1 = (z - z_0)(z - z_1) \ldots (z - z_{n-1}) $$ and the other being $$ z^n + 1 = z^n - z_0^n = (z - z_0)(z^{n-1} + z^{n-2} z_0 + \ldots + z_0^{n-1}) $$

What happens if you cancel the factors $ z - z_0 $ (using the fact that $ \mathbb C[z] $ is an integral domain) and set $ z = z_0 $?

Starfall
Starfall
September 13, 2017 14:18 PM

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