Example 3. Find the logarithmic residue of the function

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

with respect to the circle $|z|=\pi$.
Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.

In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-i, a_{2}=i$ and seven double poles

$$
z_{1}=-3, \quad z_{2}=-2, \quad z_{3}=-1, \quad z_{4}=0, \quad z_{5}=1, \quad z_{6}=2, \quad z_{7}=3
$$

Thus, $N=2$ and $P=7$. By the logarithmic residue theorem, the logarithmic residue of the function $f(z)$ with respect to the circle $|z|=\pi$ is

$$
\frac{1}{2 \pi i} \int_{|z|=\pi} \frac{f^{\prime}(z)}{f(z)} d z=2-7 \cdot 2=-12
$$

## Problems for Independent Solution

Find the logarithmic residues of the given functions with respect to the specified contours:

429. $f(z)=\frac{z}{1+z^{3}}$,

$C:|z|=2$.

430. $f(z)=\cos z+\sin z$,

$C:|z|=4$.

431. $f(z)=\left(e^{z}-2\right)^{2}$,

$C:|z|=8$.

432. $f(z)=\operatorname{th} z$,

$C:|z|=8$.

433. $f(z)=\operatorname{tg}^{3} z$,

$C:|z|=6$.

434. $f(z)=1-\operatorname{th}^{2} z$,

$C:|z|=2$.

Principle of the Argument. The logarithmic residue of the function $f(z)$ with respect to a closed contour $C$ is equal to the increment $\Delta_{C} \operatorname{Arg} f(z)$ of the argument of $f(z)$ when the contour $C$ is traversed, divided by $2 \pi$:

$$
\frac{1}{2 \pi i} \int_{C} \frac{f^{\prime}(z)}{f(z)} d z=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)
$$

Therefore, the difference between the number of zeros and poles of the function $f(z)$ enclosed in the domain $D$ is

$$
N-P=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)
$$

In other words, the difference $N-P$ is equal to the number of turns made by the vector in the $w$-plane from the point $w=0$ to the point $w=f(z)$ as the point $z$ traces the contour $C$ (the number of turns is positive if the vector rotates counterclockwise and negative otherwise).

In the special case where the function $w=f(z)$ is analytic in the domain $D$ and on its boundary $C$, where it does not vanish, the logarithmic residue of $f(z)$ with respect to $C$ gives the number of zeros of $f(z)$ in $D$, which is equal to the change in $\operatorname{Arg} f(z)$ when the contour $C$ is traversed, divided by $2 \pi$:

$$
\frac{1}{2 \pi i} \int_{C} \frac{f^{\prime}(z)}{f(z)} d z=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} f(z)=N
$$

This is the case, for example, for a polynomial $Q_{n}(z)=\sum_{k=0}^{n} a_{k} z^{k}$.