How to find an argument of this complex number? $$ z=1-e^{i\varphi}=1-\cos\varphi-i\sin\varphi $$
I know I should try to make $z$ equal to $r(\cos\psi+i\sin\psi)$. And then $Arg(z)=\psi$. But I don't really understand how to do that. I tried something like that: $$ z=2\sin^2\frac{\varphi}{2}-2i\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}=2\sin\frac{\varphi}{2}\left(\sin\frac{\varphi}{2}-i\cos\frac{\varphi}{2}\right) $$ But obviously it didn't help much.
I think you're on the right track:
\begin{align*} z&=2\sin\frac{\varphi}{2}\left(\sin \frac{\varphi}{2}-i\cos\frac{\varphi}{2}\right)\\ &=2\sin\frac{\varphi}{2}\left(-ie^{i\varphi/2}\right)\\ &=2\sin\frac{\varphi}{2}\left(e^{-i\pi/2}e^{i\varphi/2}\right)\\ &=\dots \end{align*}