<s>[INST] Complex Analysis - Laurent series of $\frac{z}{z^2+1}$ in $|z-3|>2$ I'm supposed to find the Laurent series for $\frac{z}{z^2+1}$ in $D$ : $|z-3|>2$ 
the nature of the domain $D$ tells me that the analytic part of the Laurent series expansion must be $0$
which means the series I'm looking for should look like this : 
$$\frac{a_{-1}}{z-3}+\frac{a_{-2}}{(z-3)^2}+\frac{a_{-3}}{(z-3)^3}\dots\dots$$
I tried to manipulate the function to create a  $\frac{1}{1-\frac{2}{z-3}}$ term but I keep failing. 
please help me
thanks !
 [/INST] Let $f(z)=\frac{z}{z^2+1}$.  There are pole singularities of $f$ at $z=\pm i$.  Hence, if we expand $f$ in a Laurent series around $z=3$, that series will have only a principal part for $|z-3|<\sqrt{10}$ and will have no principal part for $|z-3|>\sqrt{10}$.  
For $|z-3|>\sqrt{10}$, we can write
$$\begin{align}
\frac{z}{z^2+1}&=\frac{1/2}{z-i}+\frac{1/2}{z+i}\\\\
&=\frac{1/2}{(z-3)+3-i}+\frac{1/2}{(z-3)+3+i}\\\\
&=\frac{1}{2(z-3)}\left(\frac{1}{1+\frac{3-i}{z-3}}+\frac{1}{1+\frac{3+i}{z-3}}\right)\\\\
&=\frac{1}{2(z-3)}\sum_{n=0}^\infty (-1)^n\left(\left(\frac{3-i}{z-3}\right)^n+\left(\frac{3+i}{z-3}\right)^n\right)\\\\
&=\sum_{n=0}^\infty (-1)^n(10)^{n/2}\cos(n\arctan(1/3))(z-3)^{-(n+1)}
\end{align}$$
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