Example 4. Find the residue of the function

$$
f(z)=z^{3} \cdot \sin \frac{1}{z^{2}}
$$

at its singular point.
Solution. A singular point of the function $f(z)$ is the point $z=0$. It is an essential singular point of the function $f(z)$. Indeed, the Laurent series expansion of the function in the neighborhood of the point $z=0$ is

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

i.e., it contains an infinite number of terms in the principal part. The residue of the function at the point $z=0$ is zero, since the coefficient $c_{-1}$ in the Laurent series expansion of $f(z)$ is zero.

If the function $f(z)$ is of the form $f(z)=\frac{\varphi(z)}{\psi(z)}$, where the analytic functions $\varphi(z)$ and $\psi(z)$ have zeros of higher than the first order at the point $z_{0}$, then in this case it is convenient to replace the functions $\varphi(z)$ and $\psi(z)$ with their Taylor series expansions in the neighborhood of the point $z_{0}^{\prime}$.