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^{50}}{z_2^{10}}&=\dfrac{(\cos 15^\circ+i\sin 15^\circ)^{50}}{(\cos72^\circ+i\sin 72^\circ)^{10}}=\dfrac{\cos(15^\circ\cdot50)+i\sin(15^\circ\cdot 50)}{\cos(72^\circ\cdot10)+i\sin(72^\circ\cdot 10)}\\ &=\dfrac{\cos750^\circ+i\sin 750^\circ}{\cos720^\circ+i\sin720^\circ}=\cos(750^\circ-720^\circ)+i\sin(750^\circ-720^\circ)\\ &=\cos30^\circ+i\sin30^\circ=\dfrac{\sqrt3}{2}+\dfrac{1}{2}i \end{align*}