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_1}{z_2}&=\dfrac{\cos 75^\circ+i\sin 75^\circ}{\cos15^\circ+i\sin 15^\circ}\\ &=\cos(75^\circ-15^\circ)+i\sin(75^\circ-15^\circ)\\ &=\cos60^\circ+i\sin60^\circ \end{align*} тул $$\left(\dfrac{z_1}{z_2}\right)^3=(\cos60^\circ+i\sin60^\circ)^3=\cos(3\cdot 60^\circ)+i\sin(3\cdot 60^\circ)=-1$$