<s>[INST] How to find Laurent-series? I am trying to find the Laurent series for the function $\frac{1}{ z (2i - z)}$.
I already obtained for...
(1) ... $2  < | z |$:
$\frac{1}{ z (2i - z)} = \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i - z}   \right)
= \frac{1}{2 i} \left( \frac{1}{z} - \frac{1}{z} \cdot \frac{1}{1 - \frac{2i}{z} }  \right) 
= \frac{1}{2 i} \left[ \frac{1}{z} - \frac{1}{z} \cdot \sum_{n \geq 0} \left(  \frac{2i}{z} \right)^n  \right]    \\
= \frac{1}{2 i} \left[ \frac{1}{z}  -   \sum_{n \geq 0}  \frac{(2i)^n}{z^{n+1}}  \right]
= \frac{1}{2 i} \left[ \frac{1}{z}  -   \sum_{n \geq 0}  \frac{(2i)^n}{z^{n+1}}  \right]$ 
(2) ... $0 < | z | < 2$:
$\frac{1}{ z (2i - z)} = \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i - z}   \right)
= \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i} \cdot \frac{1}{1 - \frac{z}{2i} }  \right)   
= \frac{1}{2 i} \left( \frac{1}{z} + \frac{1}{2i} \cdot \sum_{n \geq 0} \left(  \frac{z}{2i} \right)^n   \right) 
= \frac{1}{2 i} \left( \frac{1}{z} +  \sum_{n \geq 0} \  \frac{z^n}{(2i)^{n+1}} \right)$
I am confused about the complex number $i$ in the series ... are my calculations correct?
 [/INST] Don't worry about the $i$, just treat it as you would any other complex constant. Both series you wrote look correct, but generally it is more clear to write them in the following forms.
For $|z|>2$ we can write:
$$
\frac{1}{z(2i-z)}=-\frac{1}{z^2}\left(\frac{1}{1-\frac{2i}{z}}\right)=-\frac{1}{z^2}\sum_{n=0}^{\infty}\frac{(2i)^n}{z^n}\\=\sum_{n=0}^{\infty} \frac{-(2i)^n}{z^{n+2}}
=\sum_{n=2}^{\infty} \frac{-(2i)^{n-2}}{z^n}
$$
And in the annulus $0<|z|<2$ we can similarly write:
$$
\frac{1}{z(2i-z)}=\frac{1}{2iz}\left(\frac{1}{1-\frac{z}{2i}}\right)=\frac{1}{2i}\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^n}{(2i)^n} \\
=\sum_{n=0}^{\infty} \frac{z^{n-1}}{(2i)^{n+1}} =\frac{1}{2i}\frac{1}{z}+\sum_{n=0}^{\infty} \frac{z^{n}}{(2i)^{n+2}}
$$
In particular, we can see the residue at $z=0$ is $\frac{1}{2i}$.
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