<s>[INST] 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?
 [/INST] Yes, it is correct. In fact
$$(\sqrt{3}-3i)^6=3^3(1-\sqrt{3}i)^6=2^63^3(e^{-i\pi/3})^6=1728.$$
P.S. There is no need of the term $+2\pi k$.
</s>