Example 9. Determine the nature of the singular point $z=0$ of the function

$$
f(z)=\frac{1}{2+z^{2}-2 \operatorname{ch} z}
$$
Solution. The point $z=0$ is a pole of the function $f(z)$, since it is a zero of the denominator. Consider the function $\varphi(z)=\frac{1}{f(z)}=2+z^{2}-2 \operatorname{ch} z$. For this function, $\varphi(0)=0$. Let's find the order of the zero $z=0$ of this function. We have

$$
\begin{array}{ll}
\varphi^{\prime}(z)=2 z-2 \operatorname{sh} z, & \varphi^{\prime}(0)=0 \\
\varphi^{\prime \prime}(z)=2-2 \operatorname{ch} z, & \varphi^{\prime \prime}(0)=0 \\
\varphi^{\prime \prime \prime}(z)=-2 \operatorname{sh} z, & \varphi^{\prime \prime \prime}(0)=0 \\
\varphi^{\prime \mathrm{V}}(z)=-2 \operatorname{ch} z, & \varphi^{\text {IV }}(0)=-2 \neq 0
\end{array}
$$

Thus, $z=0$ is a zero of the fourth order for $\varphi(z)$, and therefore, for the given function $f(z)$, the point $z=0$ is a pole of the fourth order.