Find the conformal mapping of the disk $\left|z\right| <2$ to the right half plane, $\operatorname{Re}(z)> 0$, such that $w(0) = 1$, $\operatorname{Arg}(w'(0)) = \frac{\pi}{2}$.
I have tried the following:
First transform radio disk $2$ to unit disk with $w_1(z)=\frac{z}{2}$ Now, I have to go from the unit circle to the right plane $\operatorname{Re}(w)> 0$ use $w_{2}(z) = - \frac{z + 1}{z-1}$.
Then the mapping that sends us from the circle of radius $2$ will be $g(z)=w_2(w_1(z))$.
Is this correct? And what do I have to do to get that $w(0) = 1$, $\operatorname{Arg}(w'(0)) = \frac{\pi}{2}$?
I would appreciate your help.
Advance:
I have to do the mapping that I need is $g(z)=-\frac{z+2}{z-2}$ I have to $g´ (0) = 1 $ Now, for me to realize that $Arg (f´(0)) = \frac{\pi}{2}$ so for it to comply, we have to take the transformation as $g (iz)$. And I conclude that the transformation that performs what I need is:
$w(z)=-\frac{iz+2}{iz-2}$
You just need a minor modification of your function. Take $w(z)=\frac {2+iz} {2-iz}$ This has all the required properties.