$(-2\sqrt{3}+2i)^{-9}$
I tried solving it like I solved other problems, I calculated $r$ which is $4$ then got $4(\sin{\frac{\pi}{6}}+\cos{\frac{\pi}{6}})$, then using an equation,
I got $4^{-9}(\sin90+\cos90)=4^{-9}(1+0)=4^{-9}$
But the answer is $\frac{i}{2^{18}}$.
I understand that $\frac{1}{4^{9}}$ is equal to $\frac{1}{2^{18}}$ but I don't understand how we got an $i$ in the answer.
Can anyone explain why?
I presume the question is to find $$(2\sqrt3+2i)^{-9}.$$ This equals $$\left(4\left(\cos\frac\pi6+i\sin\frac\pi6\right)\right)^{-9} =2^{-18}\left(\cos\left(-\frac{3\pi}2\right) +i\sin\left(-\frac{3\pi}2\right)\right)=2^{-18}i$$ via de Moivre's theorem.