3.21. Calculate

$$
\int_{L} \frac{60 e^{z}}{z(z+3)(z+4)(z+5)} d z
$$

where $L$ is the unit circle centered at the origin.
Solution. The integrand can be represented in the form

$$
\frac{f(z)}{z-a}
$$

where

$$
f(z)=\frac{60 e^{z}}{(z+3)(z+4)(z+5)}, a=0
$$

The function $f(z)$ is a single-valued analytic function in the domain $D$, such that $D=\{z \in \mathbb{C} \| z \mid \leqslant 2\}$, and $L \subset D$ is a closed continuous line bounding the region $G=\{z \in \mathbb{C} \| z \mid<1\} \subset D$. Thus, all conditions of Cauchy's integral formula are satisfied, which means:

$$
\begin{gathered}
\int_{L} \frac{60 e^{z} d z}{z(z+3)(z+4)(z+5)}=2 \pi i f(a)=2 \pi i f(0)= \\
=2 \pi i \frac{60 e^{0}}{(0+3)(0+4)(0+5)}=2 \pi i .
\end{gathered}
$$