## Task B-4.7.

Determine the modulus and argument of the complex number $z^{4}: w^{12}$, where

$$
\begin{aligned}
& z=-2 \sin \frac{5 \pi}{8}-2 i \cos \frac{13 \pi}{8} \\
& w=\sqrt{2} \cos \frac{5 \pi}{16}-i \sqrt{2} \sin \frac{5 \pi}{16}
\end{aligned}
$$
## Solution.

It holds that:

$$
\begin{aligned}
z & =-2 \cos \left(\frac{\pi}{2}-\frac{5 \pi}{8}\right)-2 i \sin \left(\frac{\pi}{2}-\frac{13 \pi}{8}\right) \\
& =-2 \cos \left(-\frac{\pi}{8}\right)-2 i \sin \left(-\frac{9 \pi}{8}\right) \\
& =-2 \cos \left(\frac{\pi}{8}\right)+2 i \sin \left(\frac{9 \pi}{8}\right) \\
& =2\left(\cos \left(\frac{9 \pi}{8}\right)+i \sin \left(\frac{9 \pi}{8}\right)\right) \\
w & =\sqrt{2}\left(\cos \frac{5 \pi}{16}-i \sin \frac{5 \pi}{16}\right) \\
& =\sqrt{2}\left(\cos \frac{27 \pi}{16}+i \sin \frac{27 \pi}{16}\right)
\end{aligned}
$$

Now we have:

$$
\begin{aligned}
z^{4}: w^{12} & =\frac{2^{4}}{(\sqrt{2})^{12}}\left(\cos \left(4 \cdot \frac{9 \pi}{8}-12 \cdot \frac{27 \pi}{16}\right)+i \sin \left(4 \cdot \frac{9 \pi}{8}-12 \cdot \frac{27 \pi}{16}\right)\right) \\
& =\frac{1}{4}\left(\cos \left(-\frac{63 \pi}{4}\right)+i \sin \left(-\frac{63 \pi}{4}\right)\right) \\
& =\frac{1}{4}\left(\cos \left(\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{4}\right)\right)
\end{aligned}
$$

The modulus of the complex number $z^{4}: w^{12}$ is $\frac{1}{4}$, and its argument is $\frac{\pi}{4}$.