Example 11. Determine the nature of the singular point $z_{0}$ of the function

$$
f(z)=\frac{1-e^{-z}}{z}
$$
Solution. Using the Taylor series expansion for the function $e^{-z}$ in the neighborhood of the point $z_{0}=0$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of zero

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

This expansion does not contain the principal part. Therefore, the point $z_{0}=0$ is a removable singularity.

The function $f(z)$, extended to the point $z=0$ by the value one, i.e.,

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-079.jpg?height=160&width=549&top_left_y=339&top_left_x=458)

is analytic at the point $z_{0}=0$.