Example 13. Determine the nature of the singular point $z=1$ of the function

$$
f(z)=(z-1) e^{1 /(z-1)}
$$
Solution. Using the expansion

$$
e^{u}=1+u+\frac{u^{2}}{2!}+\frac{u^{3}}{3!}+\ldots
$$

and setting $u=\frac{1}{z-1}$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of the point $z_{0}=1$:

$$
\begin{aligned}
f(z) & =(z-1)\left[1+\frac{1}{z-1}+\frac{1}{2!(z-1)^{2}}+\frac{1}{3!(z-1)^{3}}+\cdots\right]= \\
& =1+(z-1)+\frac{1}{2!(z-1)}+\frac{1}{3!(z-1)^{2}}+\cdots
\end{aligned}
$$

This expansion contains an infinite number of terms with negative powers of $z-1$. Therefore, the point $z_{0}=1$ is an essential singularity of the function $f(z)$.

## Problems for Independent Solution

Determine the nature of the specified singular points:

307. $\frac{1+\cos z}{z-\pi}, \quad z_{0}=\pi$.
308. $\frac{z^{2}-3 z+2}{z^{2}-2 z+1}, \quad z_{0}=1$.
309. $\frac{\sin z}{z^{2}}, \quad z_{0}=0$
310. $\frac{\operatorname{sh} z}{z}, \quad z_{0}=0$.
311. $\cos \frac{1}{z+\pi}, \quad z_{0}=-\pi$.
312. $\frac{z^{2}-1}{z^{6}+2 z^{5}+z^{4}}, \quad z_{0}=0, \quad z_{0}=-1$.
313. $\frac{\ln \left(1+z^{3}\right)}{z^{2}}, \quad z_{0}=0$.
314. $\frac{\sin ^{2} z}{z}, \quad z_{0}=0$
315. $\frac{e^{z+e}}{z+e}, \quad z_{0}=-e$.
316. $\cos \frac{1}{z}+\sin \frac{2-\pi z}{2 z}, \quad z_{0}=0$.
317. $z \operatorname{sh} \frac{1}{z}, \quad z_{0}=0$.

## § 10. Residues of Functions

Let the point $z_{0}$ be an isolated singular point of the function $f(z)$. The residue of the function $f(z)$ at the point $z_{0}$ is the number denoted by the symbol res $f\left(z_{0}\right)$ and defined by the equality

$$
\operatorname{res} f\left(z_{0}\right)=\frac{1}{2 \pi i} \oint_{\gamma} f(z) d z
$$

(Other notations: $\operatorname{res}\left[f(z) ; z_{0}\right]$, res $_{z=z_{0}} f(z)$.) The contour $\gamma$ can be taken as a circle centered at the point $z_{0}$ with a sufficiently small radius such that the circle does not extend beyond the domain of analyticity of the function $f(z)$ and does not contain other singular points of the function $f(z)$ inside it. The residue of the function is equal to the coefficient of the minus first power in the Laurent series expansion of $f(z)$ in the neighborhood of the point $z=z_{0}$:

$$
\text { res } f\left(z_{0}\right)=c_{-1} \text {. }
$$

The residue at a removable singular point is zero.

If the point $z_{0}$ is a pole of order $n$ of the function $f(z)$, then

$$
\operatorname{res} f\left(z_{0}\right)=\frac{1}{(n-1)!} \lim _{z \rightarrow z_{0}} \frac{d^{n-1}}{d z^{n-1}}\left\{f(z)\left(z-z_{0}\right)^{n}\right\}
$$

In the case of a simple pole ($n=1$)

$$
\operatorname{res} f\left(z_{0}\right)=\lim _{z \rightarrow z_{0}}\left[f(z)\left(z-z_{0}\right)\right]
$$

If the function $f(z)$ in the neighborhood of the point $z_{0}$ can be represented as the quotient of two analytic functions

$$
f(z)=\frac{\varphi(z)}{\psi(z)}
$$

where $\varphi\left(z_{0}\right) \neq 0, \psi\left(z_{0}\right)=0$, and $\psi^{\prime}\left(z_{0}\right) \neq 0$, i.e., $z_{0}$ is a simple pole of the function $f(z)$, then

$$
\operatorname{res} f\left(z_{0}\right)=\frac{\varphi\left(z_{0}\right)}{\psi^{\prime}\left(z_{0}\right)}
$$

If the point $z_{0}$ is an essential singularity of the function $f(z)$, then to find res $f\left(z_{0}\right)$, it is necessary to find the coefficient $c_{-1}$ in the Laurent series expansion of the function $f(z)$ in the neighborhood of the point $z_{0}$; this will be res $f\left(z_{0}\right)$.