Example 3. Find the zeros of the function $f(z)=1+\operatorname{ch} z$ and determine their order.
Solution.

1. Find the zeros of the analytic function $f(z)$ by solving the equation

$$
1+\operatorname{ch} z=0
$$

Since

$$
\operatorname{ch} z=\cos i z
$$

we have the equation for determining the zeros:

$$
\cos i z=-1
$$

From this, $i z=\pi(2 k+1)$ and

$$
z_{k}=\pi i(2 k+1), \quad k=0, \pm 1, \ldots
$$

2. Determine the order of each zero by computing the derivatives of the function $f(z)$:

$$
f^{\prime}(z)=\operatorname{sh} z, \quad f^{\prime \prime}(z)=\operatorname{ch} z
$$

and their values at the points $z_{k}$:

$$
f^{\prime}\left(z_{k}\right)=\operatorname{sh} \pi i(2 k+1)=0, \quad f^{\prime \prime}\left(z_{k}\right)=\operatorname{ch} \pi i(2 k+1)=-1 \neq 0
$$

Since $f\left(z_{k}\right)=f^{\prime}\left(z_{k}\right)=0$ and $f^{\prime \prime}\left(z_{k}\right) \neq 0$, then $z_{k}=i \pi(2 k+1)$ are zeros of the 2nd order of the function $f(z)=1+\operatorname{ch} z$.

Answer. $\quad z_{k}=i \pi(2 k+1) \quad(k=0, \pm 1, \ldots)$ - zeros of the 2nd order of the function $f(z)$.