3.16. Calculate

$$
\int_{L} \frac{d z}{(z-a)^{n}}
$$

where $L$ is a circle of radius $R$ centered at the point $a$, traversed counterclockwise, and $n$ is an integer.
Solution. It is clear that $L=\{z \| z-a \mid=R\}$, hence $L=\{z \mid z=$ $\left.=R e^{i \varphi}+a, \varphi \in[0,2 \pi)\right\}$. Therefore,

$$
\begin{aligned}
& \int_{L} \frac{d z}{(z-a)^{n}}=\left\{\begin{array}{c}
z=R e^{i \varphi}+a \\
d z=i R e^{i \varphi} d \varphi \\
\varphi \in[0,2 \pi)
\end{array}\right\}=\int_{0}^{2 \pi} \frac{i R e^{i \varphi} d \varphi}{\left(R e^{i \varphi}+a-a\right)^{n}}=i R \int_{0}^{2 \pi} \frac{e^{i \varphi} d \varphi}{R^{n} e^{i n \varphi}}= \\
& =i R^{1-n} \int_{0}^{2 \pi} e^{i \varphi(1-n)} d \varphi=\left[\begin{array}{l}
i \int_{0}^{2 \pi} d \varphi, n=1 ; \\
0 \\
\left.\frac{i R^{1-n}}{i(1-n)}\left\{e^{i \varphi(1-n)}\right\}\right|_{0} ^{2 \pi} ;
\end{array}=\left[\begin{array}{l}
2 \pi i, n=1 ; \\
0, n \neq 1 .
\end{array}\right.\right.
\end{aligned}
$$