Монгол Бодлогын Сан

Заавар: $1+\cos120^\circ=2\cos^260^\circ$, $\sin120^\circ=2\cos60^\circ\sin60^\circ$

Бодолт: $$\begin{align*} 1+\cos120^\circ+i\sin120^\circ&=2\cos60^\circ(\cos60^\circ+i\sin60^\circ)\\ &=2\cdot\dfrac12\cdot (\cos60^\circ+i\sin60^\circ)\\ &=\cos60^\circ+i\sin60^\circ \end{align*}$$ тул $$\begin{align*} (1+\cos120^\circ+i\sin120^\circ)^{10}&=(\cos60^\circ+i\sin60^\circ)^{10}\\ &=\cos(10\cdot60^\circ)+i\sin(10\cdot60^\circ)\}\\ &=\cos600^\circ+i\sin600^\circ\\ &=-\dfrac{1}{2}-\dfrac{\sqrt3}{2}i \end{align*}$$