Evaluate $(\sqrt{3}-3i)^6$ 
Evaluate $$(\sqrt{3}-3i)^6.$$

So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m},$$ where $m\in \mathbb{Z}.$
So $$1728e^{-2\pi i+2\pi m}=1728cos(-2\pi)=1728.$$
Is it correct?

$(\sqrt{3}-3i)^6=27(1-\sqrt{3}i)^6=27\times 2^6\Big(\dfrac{1}{2}-\dfrac{\sqrt3}{2}i\Big)^6=27\times64\space\omega^6=27\times 64$
$\omega$ is a non-real solution of the equation $x^6=1$
