Example 3. Find the order of the zero $z_{0}=0$ for the function

$$
f(z)=\frac{z^{8}}{z-\sin z}
$$
Solution. Using the Taylor series expansion of the function $\sin z$ in the neighborhood of the point $z_{0}=0$, we obtain

$$
\begin{aligned}
f(z) & =\frac{z^{8}}{z-\sin z}=\frac{z^{8}}{z-\left(z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots\right)}= \\
& =\frac{z^{8}}{\frac{z^{3}}{3!}-\frac{z^{5}}{5!}+\ldots}=\frac{z^{5}}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}=z^{5} \frac{1}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}
\end{aligned}
$$

Let

$$
\varphi(z)=\frac{1}{\frac{1}{3!}-\frac{z^{2}}{5!}+\ldots}
$$

Then $f(z)=z^{5} \varphi(z)$, where $\varphi(z)$ is a function analytic at the point $z_{0}=0$, and $\varphi(0)=6 \neq 0$. Therefore, the point $z_{0}=0$ is a zero of the fifth order for the given function.