3.23. Calculate

$$
\int_{L} \frac{e^{z}}{z^{2}+9} d z
$$

where $L$ is the circle: a) $|z-3 i|=2 ;$ b) $|z-3 i|=9$.
Solution.

a) It is obvious that:

$$
\begin{aligned}
& \int_{L} \frac{e^{z}}{z^{2}+9} d z=\frac{i}{6} \int\left(\frac{e^{z}}{z+3 i} \cdot \frac{e^{z}}{z-3 i}\right) d z= \\
= & \frac{i}{6} \int_{L} \frac{e^{z}}{z+3 i} d z-\frac{i}{6} \int_{L} \frac{e^{z}}{z-3 i} d z=\frac{i}{6}\left(I_{1}-I_{2}\right) .
\end{aligned}
$$

If the contour $L$ is the circle $|z-3 i|=2$, then the integral $I_{1}$ is zero because the function $\frac{e^{z}}{z+3 i}$ is analytic in the domain $D=\{z \in \mathbb{C}|| z-3 i \mid \leqslant 2+\delta, \delta \in[0,4)\}$, and the integral

$$
I_{2}=\int_{L} \frac{e^{z}}{z-3 i} d z
$$

equals, by Cauchy's integral formula, $I_{2}=2 \pi i e^{3 i}$.

Thus,

$$
\int_{L} \frac{e^{z}}{z^{2}+9} d z=\frac{i}{6}\left(I_{1}-I_{2}\right)=\frac{i}{6}\left(0-2 \pi i e^{3 i}\right)=-\frac{\pi}{3} e^{i(3-\pi)}
$$

b) If the contour $L$ is the circle $|z-3 i|=9$, then, according to Cauchy's integral formula, which is satisfied for the integrals $I_{1}$ and $I_{2}$, the integral $I_{1}$ equals $I_{1}=$ $=2 \pi i e^{-3 i}$, and the integral $I_{2}=2 \pi i e^{3 i}$. Thus,

$$
\begin{aligned}
\int_{L} \frac{e^{z}}{z^{2}+9} & d z=\frac{i}{6}\left(I_{1}-I_{2}\right)=\frac{i}{6}\left(2 \pi i e^{-3 i}-2 \pi i e^{3 i}\right)= \\
& =\frac{-2 \pi i^{2}}{3} \frac{e^{3 i}-e^{-3 i}}{2 i}=\frac{2 \pi \sin 3}{3}
\end{aligned}
$$