I know there are quite a few questions about Laurent series, but I couldn't find one similar to mine.
The question is as follows: Determine the Laurent series around $z=0$ of the function $\dfrac{4}{z^2+2z-3}$, which converges in $z=1+i$.
So my thoughts are, if it converges in $1+i$, it also converges everywhere in the circle with radius less than $|1+i| = \sqrt{2}$, right (forgot how that Theorem is called)? In other words, $|z| < \sqrt{2}$.
Now, write $f(z)$ as $\dfrac{4}{(z+3)(z-1)} = \dfrac{1}{z-1} - \dfrac{1}{z+3}$. So simply find the Laurent series of those two terms.
First $\dfrac{1}{z+3} = \dfrac{1}{3} \dfrac{1}{1-(-\frac{z}{3})} = \dfrac{1}{3} \displaystyle \sum_{k=0}^\infty \left( - \dfrac{z}{3} \right)^k$.
But how to rewrite $\dfrac{1}{z-1}$, considering that $|z|<\sqrt{2}$?