Calculating Residue of $f^2$ with pole of order 2

Question

The question: The function $f$ has a Pol of order 2 in $z_0$. Calculate the residue of $f^2$ in $z_0$ using the Laurentcoefficients of $f$.

My attempt: I tried to use the fact that if $f(z) = (z- z_{0})^2 g(z)$ with $g(z_0) \neq 0$ then $Res_{z_0} \frac{f'(z)}{f(z)} = k$ with that I tried to calculate the Laurentcoefficients but that does not really lead anywhere.

Greetings a Peaceful Slosh

Answer 1

Let $$ f(z)=\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\dots $$ be the Laurent series of $f$ around $z_0$. To find the reside of $f^21$ at $z_0$ we need to find the coefficient of $1/(z-z_0)$ in its Laurent series, which can be found by computing $$ f(z)^2=\Bigl(\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\dots\Bigr)^2. $$ We do not need to compute all the terms, just the $1/(z-z_0)$ one. How can $1/(z-z_0)$ appear? Only in two forms:

1. When multiplying $\dfrac{a_{-2}}{(z-z_0)^2}$ and $a_1(z-z_0)$.
2. When multiplying $\dfrac{a_{-1}}{z-z_0}$ and $a_0$. All the other terms will have other powers of $z-z_0$.