Example 5. Find the residue at the point $z=0$ of the function

$$
f(z)=\frac{\sin 3z - 3 \sin z}{(\sin z - z) \sin z}
$$
Solution. The point $z=0$ is a zero of both the numerator $\varphi(z)=\sin 3 z-3 \sin z$ and the denominator $\psi(z)=(\sin z-z) \sin z$. Let's determine the orders of these zeros using the Taylor series expansion of $\sin z$ around the point $z=0$:

$$
\sin z=z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\ldots
$$

We have

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

where

$$
\begin{aligned}
& \varphi_{1}(z)=-\frac{3^{3}-3}{3!}+\frac{3^{5}-3}{5!} z^{2}-\ldots, \varphi_{1}(0)=-4 \neq 0 \\
& \psi(z)=(\sin z-z) \sin z=\left(-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}-\cdots\right)\left(z-\frac{z^{3}}{3!}+\ldots\right)= \\
& =z^{4}\left(-\frac{1}{3!}+\frac{z^{2}}{5!}-\cdots\right)\left(1-\frac{z^{2}}{3!}+\ldots\right)=z^{4} \psi_{1}(z)
\end{aligned}
$$

where

$$
\psi_{1}(z)=\left(-\frac{1}{3!}+\frac{z^{2}}{5!}-\ldots\right)\left(1-\frac{z^{2}}{3!}+\ldots\right), \quad \psi_{1}(0)=-\frac{1}{6} \neq 0
$$

Therefore,

$$
f(z)=\frac{\sin 3 z-3 \sin z}{(\sin z-z) \sin z}=\frac{z^{3} \varphi_{1}(z)}{z^{4} \psi_{1}(z)}=\frac{\varphi_{1}(z)}{z \psi_{1}(z)}
$$

and since $\varphi_{1}(0) \neq 0, \psi_{1}(0) \neq 0$, the point $z=0$ is a simple pole of the function, so its residue at this point is found using the formula (5)

$$
\operatorname{res}_{z=0} \frac{\sin 3 z-3 \sin z}{(\sin z-z) \sin z}=\lim _{z \rightarrow 0} \frac{\varphi_{1}(z)}{z \psi_{1}(z)} z \stackrel{\psi_{-}}{=} \frac{\varphi_{1}(0)}{\psi_{1}(0)}=\frac{-4}{-\frac{1}{6}}=24
$$

## Problems for Independent Solution

Calculate the following residues:

318. $\underset{i=0}{\operatorname{res}} \frac{z^{n-1}}{\sin ^{n} z} \quad(n=1,2, \ldots)$.
319. $\operatorname{res}_{z=0} \frac{\sin 2 z-2 z}{(1-\cos z)^{2}}$.
320. $\operatorname{res}_{z=0} \frac{e^{z}-1-z}{(1-\cos 2 z) \sin z}$.
321. $\operatorname{res}_{z=0} \frac{(1-\operatorname{ch} z) \operatorname{sh} z}{(1-\cos z) \sin ^{2} z}$.
322. $\underset{x=0}{\operatorname{res}} \frac{\boldsymbol{z}^{n-2}}{\operatorname{sh}^{n} z} \quad(\boldsymbol{n}=2,3, \ldots)$.