Task B-4.3. Determine $a \in \mathbb{C}$ so that the number $z_{0}=-\sqrt{3}+i$ is a root of the polynomial $P(z)=z^{15}-a$. From the remaining roots of the polynomial $P$, determine the one with the smallest argument.
Solution. The trigonometric representation of the complex number $z_{0}=-\sqrt{3}+i$ is

$$
z_{0}=2\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)
$$

Then,

$$
z_{0}^{15}=2^{15}\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)=2^{15} i
$$

It follows that $P\left(z_{0}\right)=z_{0}^{15}-a=0$. Therefore, $a=z_{0}^{15}=2^{15} i$.

The roots of the polynomial $P$ are all the fifteenth roots of the number $a=2^{15} i=2^{15}\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$. These are the numbers

$$
w_{k}=2\left(\cos \frac{\frac{\pi}{2}+2 k \pi}{15}+i \sin \frac{\frac{\pi}{2}+2 k \pi}{15}\right), \quad k=0,1, \ldots, 14
$$

The number with the smallest argument is $w_{0}=2\left(\cos \frac{\pi}{30}+i \sin \frac{\pi}{30}\right)$.