On the residues of $f(z)=(z-z_0)^{-n}$

Question

Define $f_{n}(z)=(z-z_0)^{-n}$ where $n$ is a positive integer. Notice that $f_n$ has a pole of order $n$ at $z=z_0$, for every positive integer $n$.

My question: Does $f$ has residue $0$ at $z=z_0$ for every positive integer $n$?

Unless i'm mistaken, this seems to be the case, since

$\frac{1}{(n-1)!}\frac{d^{n} f}{dz^{n}} (z-z_0)^n (z-z_0)^{-n}=0$

for every positive integer $n.$

Answer 1

No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should be $\frac{\mathrm d^{n-1}f}{\mathrm dx^{n-1}}$ instead of $\frac{\mathrm d^nf}{\mathrm dx^n}$.

Answer 2

The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form

$$g(z):=\sum_{k\in\Bbb Z} c_k (z-z_0)^k$$

In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $k\neq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $n\neq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.