3.33. Compute

$$
\int \frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}} d z
$$

where $L$ is the circle $|z|=1$.
Solution. A special point of the integrand function $f(z)=\frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}}$ is the point $z=-\frac{\pi i}{2}$, which represents a pole of second order. There are no other special points of the function $f(z)$ within the region bounded by the circle $|z|=1$. Therefore, by the main theorem of residues, we have:

$$
\int_{L} \frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}} d z=2 \pi i \operatorname{res}\left(\frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}},-\frac{\pi i}{2}\right)
$$

By the general formula for computing the residue at a pole of order $n$, and in this case for $n=2$, we have:

$$
\begin{gathered}
\operatorname{res}\left(\frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}},-\frac{\pi i}{2}\right)=\frac{1}{(2-1)!} \lim _{z \rightarrow-\frac{\pi i}{2}} \frac{d}{d z}\left[\left(z+\frac{\pi i}{2}\right)^{2} \cdot \frac{e^{2 z}}{\left(z+\frac{\pi i}{2}\right)^{2}}\right]= \\
=\lim _{z \rightarrow-\frac{\pi i}{2}} \frac{d}{d z} e^{2 z}=\lim _{z \rightarrow-\frac{\pi i}{2}} e^{2 z}=e^{-\frac{\pi i}{2}}=\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}=-i
\end{gathered}
$$

Therefore,

$$
\int \frac{e^{2 z}}{\left(z+\frac{\pi l}{2}\right)^{2}} d z=2 \pi i \cdot(-i)=2 \pi
$$