Showing that $i\arg(e^{2z})=2iy?$ Does $i\arg(e^{2z})=2iy?$ If it does I have solved my problem, and hence it seems like it must  be the case, but I don't see it.
$$i\arg(e^{2z})=i\arg(e^{2x+2iy})=i\arg(e^{2x}e^{2iy})\implies \theta=2y(?)$$ Why does the $2x$ get 'ignored'?

A property of the argument is $\arg(z_1z_2) = \arg(z_1)+\arg(z_2)$. Therefore, 
$$
\arg(e^{2x}e^{2iy}) = \arg(e^{2x})+\arg(e^{2iy})
$$
For $\arg\in(-\pi,\pi)$, what is $ \arg(e^{2x})$?
