![\bf \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ z=\stackrel{a}{-1}\stackrel{b}{-\sqrt{3}~i}~~ \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right)\\[-0.5em] \hrulefill\\ r=\sqrt{(-1)^2+(\sqrt{3})^2}\\ \qquad \sqrt{4}\\ \qquad 2\\ \theta =tan^{-1}\left( \frac{-\sqrt{3}}{-1} \right)\\\\ \qquad \frac{4\pi }{3} \end{cases}](/tpl/images/0298/8173/b6186.png)
![\bf z=2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right]\implies z^6=\left[ 2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right] \right]^6 \\\\\\ z^6=2^6\left[ cos\left( 6\cdot \frac{4\pi }{3} \right) -i~sin\left( 6\cdot \frac{4\pi }{3} \right)\right]\implies z^6=64[cos(8\pi )-i~sin(8\pi )] \\\\\\ z^6=64[(-1)-i~(0)]\implies z^6=\stackrel{\stackrel{a}{\downarrow }}{-64}~~\stackrel{\stackrel{b}{\downarrow }}{-0i}](/tpl/images/0298/8173/d0aa8.png)
Ineed plz. its demoivres theorem
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