3.29. Expand the function $f(z)=\frac{2 i a}{z^{2}+a^{2}}$ into a Laurent series in the annulus $0<|z-i a|<a$, where $a$ is a positive real number.
Solution. In the ring $0<|z-i a|<a$, the function $f(z)$ is analytic, we can expand it into a Laurent series:

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