Could anyone help me to think about a conformal map from $\{0<Re(z)<\frac{\pi}{2}\}$ to $\{0<Im(z)<\pi\}$?
And how could we approach the question about finding a conformal mapping? I know that $z^2$ doubles the angle, but are there any other conformal mapping widely used?
Thank you!
I'd say $f: z \mapsto 2iz$. The reason behind that is that the multiplication by $2$ will have the effect of transforming $(0,\frac{\pi}{2})$ into $(0,\pi)$, and will transform $Re(z)$ into $Im(z)$.
Proof: if $Re(z) \in (0,\frac{\pi}{2})$, we can write: $z=a+ib$ with $a \in (0,\frac{\pi}{2})$. Then $f(z)=2iz = -2b+2ia$, therefore $Im(f(z)) = 2a \in (0,\pi)$. It is obvious that $f$ is bijective.
It is obvious that $f$ is conform because it is the composite of an homothety (multiplication by $2$) and a rotation (multiplication by $i$), so the angle between two vectors is the same angle between their images.