Example 6. Find the residues of the function

$$
f(z)=\frac{e^{1 / z}}{1-z}
$$

at its singular points.
Solution. The singular points of the given function are $z=1$ and $z=0$. The point $z=1-$ is a simple pole, therefore

$$
\operatorname{res}_{z=1} f(z)=\left.\frac{e^{1 / z}}{-1}\right|_{z=1}=-e
$$

To determine the nature of the singular point $z=0$, we expand the function into a Laurent series in the neighborhood of this point. We have

$$
e^{1 / z}=1+\frac{1}{z}+\frac{1}{2!z^{2}}+\frac{1}{3!z^{3}}+\ldots, \quad \frac{1}{1-z}=1+z+z^{2}+z^{3}+\ldots
$$

Multiplying these series, we get

$$
\begin{aligned}
\frac{e^{1 / z}}{1-z} & =\left(1+\frac{1}{z}+\frac{1}{2!z^{2}}+\frac{1}{3!z^{3}}+\cdots\right)\left(1+z+z^{2}+z^{3}+\ldots\right)= \\
& =\left(1+\frac{1}{2!}+\frac{1}{3!}+\ldots\right) \frac{1}{z}+c_{-2} \frac{1}{z^{2}}+\ldots+\text { regular part }
\end{aligned}
$$

where $c_{-k} \neq 0, k=2,3, \ldots$.

Since the principal part of the Laurent series contains an infinite number of terms with negative powers of $z$, the point $z=0$ is an essential singular point of the given function. Its residue at $z=0$ is

$$
\text { res } f(z)=c_{-1}=1+\frac{1}{2!}+\frac{1}{3!}+\ldots=e-1
$$