3.30. Expand the function $f(z)=\frac{z+2}{z^{2}+4 z+3}$ into a Laurent series in the annulus $2<|z+1|<+\infty$.
Solution. The function $f(z)$ has two isolated singular points, namely $z=-1$ and $z=-3$. In the annulus $2<|z+1|<+\infty$, the function $f(z)$ is analytic, and we can expand it into a Laurent series:

$$
\begin{aligned}
f(z)= & \frac{z+2}{z^{2}+4 z+3}=\frac{z+2}{(z+3)(z+1)}=\frac{z+2}{(z+1)\left(1+\frac{2}{z+1}\right)}= \\
& =\frac{z+2}{z+1} \sum_{k=0}^{\infty} \frac{2^{k}}{(z+1)^{k}}=(z+2) \sum_{k=0}^{\infty} \frac{2^{k}}{(z+1)^{k+1}}
\end{aligned}
$$