<s>[INST] Finding the Laurent representation of a complex function How can i find the Laurent representation fot the function:
$$f(z)=\dfrac{1}{1-z^2}+\dfrac{1}{3-z}$$
In the region of:
a) $\{z\in\mathbb C:1<|z|<3\}$
b) $\{z\in\mathbb C:1<|z-2|<3\}$
 [/INST] $\frac{1}{1 - z} = \sum z^n$--the simplest function.  You have the following:
$$
\frac{1}{2}\frac{1}{1 - (-z)} + \frac{1}{2}\frac{1}{1 - z} + \frac{1}{3}\frac{1}{1 - \frac{z}{3}}
$$
When $1 < |z| < 3$, we need the Laurent series for $\frac{1}{1 \pm z}$ but the Taylor series for $\frac{1}{3 - z}$:
$$
\frac{1}{1 - z} = \frac{1}{z}\frac{1}{z^{-1} - 1} = -z^{-1}\frac{1}{1 - z^{-1}} = -z^{-1}\sum_0^\infty z^{-n} = -\sum_1^\infty z^{-n}
$$
Which gives:
\begin{align}
\frac{1}{1 - z} =&\ -\sum_1^\infty z^{-n} \\
\frac{1}{1 - (-z)} =&\ -\sum_1^\infty (-1)^n * z^{-n} \\
\frac{1}{1 - \frac{z}{3}} =&\ \sum_0^\infty \left(\frac{z}{3}\right)^n
\end{align}
Adding this gives something to the affect of:
$$
a_n = \begin{cases}
\frac{1}{3} & n = 0 \\
-z^{-n} + \frac{1}{3}\frac{z^n}{3^n} & n \text{ even} \\
\frac{1}{3}\frac{z^n}{3^n} & n \text{ odd}
\end{cases}
$$
I'm not quite clear on the domain of the second part.
p.s. You actually don't need the partial fraction decomposition (and I believe there was a previous answer which did not use partial fractions).
We know that $\frac{1}{1 - z^2}$'s Taylor series has a radius of convergencce of $|z| < 1$ (and is clearly undefined for $z = \pm 1$), and finally since the Laurent series for $\frac{1}{1 - z} = -\sum_1^\infty z^{-n}$, meaning:
$$
\frac{1}{1 - z^2} = -\sum_1^\infty z^{-2n} \text{, for } |z| > 1
$$
Adding to the Taylor series we get:
$$
-\sum_1^\infty z^{-2n} + \frac{1}{3}\sum_0^\infty \left(\frac{z}{3}\right)^n
$$
...but the exponents don't match up (which would be nice)...so match them up by splitting the Taylor series into two sums:
$$
-\sum_1^\infty z^{-2n} + \frac{1}{3}\left(\sum_0^\infty \left(\frac{z}{3}\right)^{2n} + \sum_0^\infty \left(\frac{z}{3}\right)^{2n + 1} \right)
$$
Which can be written as:
$$
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_0^\infty\left(\frac{z}{3}\right)^{2n + 1}\\
\text{or} \\
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_1^\infty\left(\frac{z}{3}\right)^{2n - 1} \\
\text{or}\\
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n} + \frac{1}{3}\left(\frac{z}{3}\right)^{2n - 1}\right)
$$
</s>