Example 5. Find the zeros of the function

$$
f(z)=\frac{\sin ^{4} z}{z^{2}}
$$

and determine their order.
Solution.

1. Function

$$
f(z)=\frac{\sin ^{4} z}{z^{2}}
$$

is analytic for all $z \neq 0$ (the function $f(z)$ is not defined at $z=0$). We find the zeros of the function by solving the equation $f(z)=0$ (for $z \neq 0$).

We obtain

$$
z_{k}=\pi k, \quad k= \pm 1, \pm 2, \pm 3, \ldots
$$

The points $z_{k}=\pi k$ are simple zeros of the function $\sin z$ and, consequently, are zeros of the 4th order of the function $\sin ^{4} z$. Since $z^{2} \neq 0$ when $z=\pi k$, we have

$$
\frac{\sin ^{4} z}{z^{2}}=\left(z-z_{k}\right)^{4} \cdot \varphi(z), \quad \varphi\left(z_{k}\right) \neq 0
$$

From this, we conclude that the points

$$
z_{k}=\pi k, \quad k= \pm 1, \pm 2, \pm 3, \ldots
$$

are zeros of the 4th order of the function $f(z)$.

Answer. $z_{k}=\pi k, (k= \pm 1, \pm 2, \pm 3, \ldots)$ are zeros of the 4th order of the function $f(z)$.

Conditions of the problem. Find the zeros of the analytic function $f(z)$ and determine their order.

1. $f(z)=\sin z-z$. 2. $\left(z^{2}+1\right) \operatorname{sh} \pi z$. 3. $f(z)=1-\cos z-z^{2} / 2$. 4. $(z+1)^{2} \sin \pi z$. 5. $f(z)=\left(e^{2 z}-1-2 z\right) \sin z$. 6. $f(z)=z(1-\operatorname{ch} z)$. 7. $f(z)=(z-\operatorname{sh} z)^{2}$. 8. $f(z)=\sin ^{2} z / z^{3}$. 9. $f(z)=z \operatorname{tg} z$. 10. $f(z)=\operatorname{sh}^{2} z / z$.

Answers. 1. $z=0$ - zero of the 3rd order. 2. $z= \pm i$ - zeros of the 2nd order. 3. $z=2 \pi k(k=0, \pm 1, \ldots)$ - zeros of the 4th order. 4. $z=-1$ - zero of the 3rd order; $z=k(k= \pm 1, \pm 2, \ldots)$ - zeros of the 1st order. 5. $z=0$ - zero of the 3rd order; $z=\pi k(k= \pm 1, \pm 2, \ldots)$ - zeros of the 1st order. 6. $z=0$ - zero of the 2nd order; $z=2 \pi k i(k= \pm 1, \pm 2, \ldots)$ - zeros of the 2nd order. 7. $z=0$ - zero of the 6th order. 8. $z=\pi k$ $(k= \pm 1, \pm 2, \ldots)$ - zeros of the 2nd order. 9. $z=0$ - zero of the 2nd order; $z=\pi k(k= \pm 1, \pm 2, \ldots)$ - zeros of the 1st order. 10. $z=\pi k i$ $(k= \pm 1, \pm 2, \ldots)$ - zeros of the 2nd order.

### 1.13. Type of Isolated Singular Point

Problem Statement. Determine the type of isolated singular point $z=a$ of the function $f(z)$.

Solution Plan. Since $z=a$ is an isolated singular point of the function $f(z)$, there exists a neighborhood of this point where $f(z)$ can be expanded into a Laurent series.

1. Find the Laurent series expansion of the function $f(z)$ in the neighborhood of the point $z=a$:

$$
f(z)=\sum_{n=1}^{\infty} \frac{c_{-n}}{(z-a)^{n}}+\sum_{n=0}^{\infty} c_{n}(z-a)^{n}
$$

The first sum in (1) is called the principal part of the Laurent series.

2. Determine the number of terms in the principal part of the Laurent series of the function $f(z)$.

If the principal part is absent (all $c_{-n}=0$), then $z=a$ is a removable singular point of the function $f(z)$.

If the principal part contains a finite number of terms, then $z=a$ is a pole, and its order is equal to the highest power of $1 /(z-a)$.

If the principal part contains an infinite number of terms, then $z=a$ is an essential singular point of the function $f(z)$.

Remark. The type of singular point of the function $f(z)$ can also be determined by computing the limit of $f(z)$ as $z \rightarrow a$:

if $\lim _{z \rightarrow a} f(z)=A$, then $z=a$ is a removable singular point;

if $\lim _{z \rightarrow a} f(z)=\infty$, then $z=a$ is a pole ${ }^{*)}$;

if $\lim _{z \rightarrow a} f(z)$ does not exist, then $z=a$ is an essential singular point.