Монгол Бодлогын Сан

Заавар: $$\dfrac{\cos\alpha+i\sin\alpha}{\cos\beta+i\sin\beta}=\cos(\alpha-\beta)+i\sin(\alpha-\beta)$$ $$(\cos\alpha+i\sin\beta)^n=\cos n\alpha+i\sin n\alpha$$

Бодолт: $$\begin{align*} \dfrac{z_2}{z_1}&=\dfrac{\cos 15^\circ-i\sin 15^\circ}{\cos75^\circ-i\sin 75^\circ}\\ &=\dfrac{\cos (-15^\circ)+i\sin(-15^\circ)}{\cos(-75^\circ)+i\sin(-75^\circ)}\\ &=\cos\big(-15^\circ-(-75^\circ)\big)+i\sin\big(-15^\circ-(-75^\circ)\big)\\ &=\cos60^\circ+i\sin60^\circ \end{align*}$$ тул $$\left(\dfrac{z_2}{z_1}\right)^3=(\cos60^\circ+i\sin60^\circ)^3=\cos(3\cdot 60^\circ)+i\sin(3\cdot 60^\circ)=-1$$