# Finding the Laurent series of $\dfrac{4}{z^2+2z-3}$

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 $|z|<\sqrt{2}$?

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You have poles at $1$ and $-3$. You know that Laurent series converge in annuli. You're looking for the Laurent series that converges in $1 + i$ which is inside the annulus $\{1 \lt |z| \lt 3\}$, so ... – t.b. Nov 13 '11 at 18:41
Thanks, for some reason I entirely forgot about the annulus, but considered a disc (since it is stated that the Laurent series must be determined around $z=0$). I'll have to review this subject some more in my book. – Ailurus Nov 13 '11 at 19:29

The function $f(z)=4/(z^2+2\,z-3)$ has two singularities, both poles of order one: $z=1$ and $z=-3$. There are then three regions to take into account when computing the Laurent series of $f$ around $z=0$: $D_1=\{\,|z|<1\,\}$, $D_2=\{\,1<|z|<3\,\}$ and $D_3=\{\,|z|>3\,\}$. All you have to do is finf in which of them is $1+i$, and find the Laurent series that converges in that region.
Ah I mixed up some things, now it is clear. So I can still use my expression for $\dfrac{1}{z+3}$, and for $\dfrac{1}{z-1}$ I can use $\dfrac{1}{z} \; \dfrac{1}{1- {1 \over z}}$. By the way, how would this story change if I would like to determine the Laurent series around something else than $z=0$? – Ailurus Nov 13 '11 at 19:26
Let's say you want the Laurent series around $z=z_0$ (with the same $f$). Then you would have again three regions (unless $z_0$ is one of the poles), determined by the distance of $z_0$ to $1$ and $-3$. One possibility fir doing the computations is to make the change of variable $\zeta=z-z_0$ and find the Laurent series around $\zeta=0$ of $g(\zeta)=f(\zeta+z_0)$. – Julián Aguirre Nov 13 '11 at 19:51