:$$\begin{align*}
z^n=&\sin\left(n~\mathrm{Arg}~z^n\right)+i\sin\left(n~\mathrm{Arg}~z^n\right)\\\overline{z}^{~n}=&\sin\left(n~\mathrm{Arg}~z^n\right)-i\sin\left(n~\mathrm{Arg}~z^n\right)\\z^n-\overline{z}^{~n}=&2i\sin\left(n~\mathrm{Arg}~z^n\right)
\end{align*}$$
