<s>[INST] Classification of singularity and Laurent series We have the following complex function:
$$f(z) = \frac{1}{z^2+4}$$
We need to get the classification of singularities and also the main part (The one with negative coefficients) of Laurent series around the point $2i$.
What we have done is the following:
$$z^2 +4 = (z+2i)(z-2i) \implies h \,\text{has a pole in 2i of degree 1 }$$
$$m = 1 \implies c_{-1} = \lim_{z\rightarrow 2i}\frac{1}{z+2i} = -\frac{i}{4} \implies $$
The main part of Laurent is $-\frac{i}{4}(z-2i)^{-1}$
However I don't really understand the procedure above. And usually when we solved problems like that we did something like $w = z-3$ if we would do it around the point $3$. Why don't we use $w = z-2i$ here ?
 [/INST] This technique is a convenient shortcut to determine the residuum of a Laurentseries expansion at a simple pole $z_0$. Doing it the long way we obtain
\begin{align*}
\color{blue}{f(z)=\frac{1}{z^2+4}}&=\frac{1}{(z-2i)(z+2i)}\tag{1}\\
&=\frac{1}{z-2i}\,\frac{1}{4i+(z-2i)}\\
&=\frac{1}{z-2i}\,\frac{1}{4i\left(1+\frac{z-2i}{4}\right)}\\
&\color{blue}{=\frac{1}{z-2i}\,\frac{1}{4i}}\left(1-\frac{z-2i}{4i}+\left(\frac{z-2i}{4i}\right)^2-\cdots\right)\tag{2}
\end{align*}
We determine from (2) the residuum of $f$ at $z=2i$ and get
\begin{align*}
\color{blue}{\lim_{z\to 2i}f(z)(z-2i)}=\frac{1}{4i}\left(1-0+0-0+\cdots\right)\color{blue}{=-\frac{i}{4}}
\end{align*}
Since $z_0$ is a simple pole, the main part consists of just one term and we have
\begin{align*}
\color{blue}{-\frac{i}{4}\frac{1}{z-2i}}
\end{align*}

Comparing (1) with (2) we observe we can also calculate the residuum by using as shortcut the representation (1) and obtain
\begin{align*}
\color{blue}{\lim_{z\to 2i}f(z)(z-2i)}=\lim_{z\to 2i}\frac{1}{z+2i}=\frac{1}{4i}\color{blue}{=-\frac{i}{4}}
\end{align*}

</s>