3. Zadatak

Množenjem jednadžbe sa $z$ dobivamo

$$z^2-2\sin\frac\pi7\,z+1=0.$$

Rješenja su

$$\begin{aligned} z_{1,2} &=\frac{2\sin\frac\pi7\pm\sqrt{4\sin^2\frac\pi7-4}}2\\ &=\frac{2\sin\frac\pi7\pm2\sqrt{-(1-\sin^2\frac\pi7)}}2\\ &=\sin\frac\pi7\pm i\cos\frac\pi7. \end{aligned}$$

Trigonometrijski zapis dobivenih rješenja je

$$z_1=\cos\left(\frac\pi2-\frac\pi7\right) +i\sin\left(\frac\pi2-\frac\pi7\right) =\cos\frac{5\pi}{14}+i\sin\frac{5\pi}{14},$$ $$z_2=\cos\frac{5\pi}{14}-i\sin\frac{5\pi}{14} =\cos\frac{23\pi}{14}+i\sin\frac{23\pi}{14}.$$

Prema tome,

$$z_1^{770} =\cos(275\pi)+i\sin(275\pi)=-1,$$

a

$$z_2^{770} =\cos(1265\pi)+i\sin(1265\pi)=-1.$$

Zato je

$$z_1^{770}+z_2^{770}=-2.$$