1. Zadatak

Uvodimo zamjenu $z^2=t$. Tada dana jednadžba prelazi u $$t^2+2\cdot(1-i)\cdot t-2i=0$$ $$t_{1,2}=\frac{-2(1-i)\pm\sqrt{4\cdot(-2i)+8i}}2=\frac{-2+2i}2$$ $$z^2=-1+i$$ $$z=\sqrt{-1+i},\qquad -1+i=\sqrt2\left(\cos\frac{3\pi}4+i\sin\frac{3\pi}4\right)$$ $$z_1=\sqrt[4]2\left(\cos\frac{3\pi}8+i\sin\frac{3\pi}8\right),\qquad z_2=\sqrt[4]2\left(\cos\frac{11\pi}8+i\sin\frac{11\pi}8\right)$$