Z=-1+i√3. Найдите z^9

Решите задачу:

$z=-1+i\sqrt3; \ \ \ z^9-?\\ z=-1+\sqrt{3}i=re^{i\phi};\\ r=\sqrt{\left(-1\right)^2+\left(\sqrt{3}\right)^2}=\sqrt{1+3}=\sqrt{4}=2;\\ \phi=\arctan\left(\frac{\sqrt{3}}{-1}\right)=\arctan\left(-\sqrt{3}\right)=-\arctan\left(\sqrt3\right)=\\ =-\frac{2\pi}{3}+\pi n;\\ z=2\cdot e^{\frac{2\pi}{3}};\\ z^9=2^9\cdot e^{\frac{18\pi}{3}+9\pi}=512\cdot e^{6\pi}=512\left(\cos\left(6\pi\right)+i\sin\right\left(6\pi\right))=512\cdot\left(1+0\right)=\\ =512.$