Example 13. Expand the function

$$
f(z)=z^{2} \cos \frac{1}{z}
$$

into a Laurent series in the neighborhood of the point $z_{0}=0$.
Solution. For any complex $\zeta$ we have

$$
\cos \zeta=-\frac{\zeta^{2}}{2!}+\frac{\zeta^{4}}{4!}-\frac{\zeta^{6}}{6!}+\cdots
$$

Setting $\zeta=\frac{1}{z}$, we get

$$
z^{2} \cos \frac{1}{z}=z^{2}\left(1-\frac{1}{2!z^{2}}+\frac{1}{4 \cdot z^{4}} \frac{1}{6!z^{6}}+\ldots\right)=z^{2}-\frac{1}{2!}+\frac{1}{4 \cdot z^{2}} \frac{1}{6!z^{4}}+\cdots
$$

or

$$
z^{2} \cos \frac{1}{z}=-\frac{1}{2}+z^{2}+\frac{1}{4!z^{2}}-\frac{1}{6!z^{4}}+\ldots
$$

This expansion is valid for any point $z \neq 0$. In this case, the "ring" represents the entire complex plane with one removed point $z=0$. This "ring" can be defined by the following relation: $0<|z-0|<+\infty$. Here $r=0, \boldsymbol{R}=+\infty, z_{0}=0$. The given function is analytic in the specified "ring".