<s>[INST] Finding Laurent series of $f(z) = \frac{1}{z^{2}+4}$ on two different domains. I need to find the Laurent series of $$f(z) = \frac{1}{z^{2}+4}.$$ First for $ z \in \mathbb{C}: |z|<2$ and then for $z \in \mathbb{C}: 1<|z-i|<3$.
Now for the first restriction I do the following:
\begin{align}
f(z) &= \frac{1}{4} \frac{1}{1+\frac{z^{2}}{4}}\\
&= \frac{1}{4} \sum\limits_{n=0}^{\infty} \left(-\frac{z^{2}}{4} \right)^{n}\\
&= \frac{1}{4} \sum\limits_{n=0}^{\infty} (-1)^{n} \left(\frac{1}{4}\right)^{n} z^{2n}\\
&= \frac{1}{4} \left( 1 -\frac{z^{2}}{4} + \frac{z^{4}}{4^{2}} -\frac{z^{6}}{4^{3}} + \dots       \right)\\
&= \left( \frac{1}{4} -\frac{z^{2}}{4^{2}} + \frac{z^{4}}{4^{3}} -\frac{z^{6}}{4^{4}} + \dots       \right),
\end{align}
which I think is all right and I used the fact $\frac{1}{1-z} = \sum\limits_{n=0}^{\infty} z^{n}$ which is valid for $|z|<1$ (which makes the above valid since $\frac{-z^{2}}{4}<1$ implies $|z|<2$.
So basically my first question is if the above is valid and the second question is how to proceed for $1<|z-i|<3$?
 [/INST] You can write
\begin{align*}
\frac{1}{z^2+4} & =\frac{1}{4i}\left[\frac{1}{z-2i}-\frac{1}{z+2i}\right]\\
& =\frac{1}{4i}\left[\frac{1}{(z-i)-i}-\frac{1}{(z-i)+3i}\right]\\
& =\frac{1}{4(z-i)}\left[\frac{1}{1-\left(\frac{i}{z-i}\right)}\right]+\frac{1}{12}\left[\frac{1}{1+\left(\frac{z-i}{3i}\right)}\right]\\
&=\frac{1}{4(z-i)} \sum_{k=0}^\infty\left(\frac{i}{z-i}\right)^k+\frac{1}{12}\sum_{k=0}^\infty(-1)^k\left(\frac{z-i}{3i}\right)^k
\end{align*}
Note: for the first expression, the series is valid for $\left|\frac{i}{z-i}\right|<1 \implies 1 < |z-i|$ and for the second expression, the series is valid for $\left|\frac{z-i}{3i}\right|<1 \implies |z-i|<3$.
Can you take it from here?
</s>