Example 12. Determine the nature of the singular point $z_{0}=0$ of the function

$$
f(z)=\frac{1-\cos z}{z^{7}}
$$
Solution. Expanding the function $\cos z$ into a Taylor series in powers of $z$, we obtain the Laurent series expansion of the function $f(z)$ in the neighborhood of zero:

$$
\begin{aligned}
f(z) & =\frac{1}{z^{7}}\left(\frac{z^{2}}{2!}-\frac{z^{4}}{4!}+\frac{z^{6}}{6!}-\frac{z^{8}}{8!}+\frac{z^{10}}{10!}-\cdots\right)= \\
& =\frac{1}{2!z^{5}}-\frac{1}{4!z^{3}}+\frac{1}{6!z}-\frac{z}{8!}+\frac{z^{3}}{10!}-\cdots
\end{aligned}
$$

The Laurent series expansion of the function $f(z)$ in the neighborhood of the point $z_{0}=0$ contains a finite number of terms with negative powers of $z$. Therefore, the point $z_{0}=0$ is a pole of order five, since the highest power of $z$ in the denominators of the terms in the principal part of the Laurent series is five.