Let $$z= 1- i \sqrt 3$$ Find polar representation of $$iz^-2$$ I have a solution but the solution given me seems to be different from what I get. What I got: $$1/4 e^{-i5/6 \pi}$$ while the solution gives $$2 e^{i1/6 \pi}$$
Not sure how I went wrong, any advice?
EDIT: My workings.
I'm not sure which part of my working went wrong, so I'm just going to list out my workings:
$$\begin{aligned}z=1-i\sqrt {3}\\ \left| r\right| =\sqrt {4}=2\\ \arg \left( z\right) =-\dfrac {\pi }{3}\\ z=2e^{i\dfrac {\pi }{3}}\\ i=e^{i\dfrac {\pi }{2}}\\ iz^{-2}=\dfrac {i}{z^{2}}\end{aligned}$$ $$\dfrac {i}{z^{2}}=\dfrac {1}{2^{2}}\left[ ( \cos \left( \dfrac {\pi }{2}-2\left( \dfrac {\pi }{3}\right) \right) +i\sin ( \dfrac {\pi }{2}-2\left( \dfrac {\pi }{3}\right) \right]$$
That's how I got my argument and new angle.