Example 3. Find the residues of the function $f(z)=\frac{1}{z^{4}+1}$ at its singular points.
Solution. The singular points of $f(z)$ are the zeros of the denominator, i.e., the roots of the equation $z^{4}+1=0$. We have

$$
z_{1}=e^{i \pi / 4}, \quad z_{2}=e^{i 3 \pi / 4}, \quad z_{3}=e^{-i 3 \pi / 4}, \quad z_{4}=e^{-i \pi / 4}
$$

Using formula (5), we get

$$
\begin{aligned}
& \operatorname{res} f\left(z_{1}\right)=\left.\frac{1}{4 z^{3}}\right|_{z=e^{i \pi / 4}}=\frac{1}{4} e^{-i 3 \pi / 4} \\
& \operatorname{res} f\left(z_{2}\right)=\left.\frac{1}{4 z^{3}}\right|_{z=e^{i 3 \pi / 4}}=\frac{1}{4} e^{-i 9 \pi / 4} \\
& \operatorname{res} f\left(z_{3}\right)=\left.\frac{1}{4 z^{3}}\right|_{z=e^{-i 3 \pi / 4}}=\frac{1}{4} e^{i 9 \pi / 4} \\
& \operatorname{res} f\left(z_{4}\right)=\left.\frac{1}{4 z^{3}}\right|_{z=e^{-i \pi / 4}}=\frac{1}{4} e^{i 3 \pi / 4}
\end{aligned}
$$