3.22. Calculate

$$
\int_{L} \frac{\sin 3 z}{(6 z-\pi)^{3}} d z
$$

where $L$ is the unit circle centered at the origin.
Solution. The integrand can be represented in the form

$$
\frac{f(z)}{(z-a)^{3}}
$$

where $f(z)=6^{-3} \sin 3 z, a=\frac{\pi}{6}$. The function $f(z)$ is a single-valued analytic function in the domain $D$, such that

$$
D=\left\{z \in \simeq \| z \left\lvert\, \leqslant \frac{\pi}{2}\right.\right\}
$$

and $L \subset D$ is a closed continuous line bounding the region $G=\{z \in C \| z \mid<1\} \subset D$. Thus, all conditions of Cauchy's integral formula are satisfied, and by its corollary, we have:

$$
\int_{L} \frac{\sin 3 z}{(6 z-\pi)^{3}} d z=\frac{2 \pi i}{2!} f^{n}(a)=i \pi f^{n}\left(\frac{\pi}{6}\right)=-\frac{i \pi}{24}
$$