1. Zadatak

Prvo rješenje

Zapišimo prvo zadane brojeve u trigonometrijskom obliku:

$$\begin{aligned} z&=-\cos\frac{2\pi}{5}-i\sin\frac{3\pi}{5}\\ &=\cos\left(\pi+\frac{2\pi}{5}\right)+i\sin\left(2\pi-\frac{3\pi}{5}\right)\\ &=\cos\frac{7\pi}{5}+i\sin\frac{7\pi}{5}, \end{aligned}$$ $$w=1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right).$$

Zaključujemo da je

$$z\cdot w =\sqrt2\left(\cos\left(\frac{7\pi}{5}+\frac{\pi}{4}\right) +i\sin\left(\frac{7\pi}{5}+\frac{\pi}{4}\right)\right) =\sqrt2\left(\cos\frac{33\pi}{20}+i\sin\frac{33\pi}{20}\right),$$

odnosno

$$\begin{aligned} (z\cdot w)^{20} &=(\sqrt2)^{20}\left(\cos\frac{20\cdot33\pi}{20} +i\sin\frac{20\cdot33\pi}{20}\right)\\ &=2^{10}(\cos33\pi+i\sin33\pi)=-1024. \end{aligned}$$

Drugo rješenje

Uočimo da vrijedi

$$z=-\cos\frac{2\pi}{5}-i\sin\frac{3\pi}{5} =-\cos\frac{2\pi}{5}-i\sin\frac{2\pi}{5} =-\left(\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\right),$$ $$w^2=(1+i)^2=2i.$$

Slijedi

$$\begin{aligned} z^{20} &=\left(-\left(\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\right)\right)^{20}\\ &=\cos\frac{20\cdot2\pi}{5}+i\sin\frac{20\cdot2\pi}{5}\\ &=\cos8\pi+i\sin8\pi=1, \end{aligned}$$ $$w^{20}=(w^2)^{10}=(2i)^{10}=1024i^2=-1024.$$

Konačno je

$$(z\cdot w)^{20}=z^{20}\cdot w^{20}=-1024.$$