I want to find a conformal map $f$ between $D = \{z\in\Bbb C : |Arg(z)|<\frac \pi4\}\setminus[1,+\infty)$ and $D(0,R)$, for $R$ a real positive number that I have to compute as well. It also has to verify that, for $z_0 = \frac 12$, $f(z_0) = 0$ and $f'(z_0) = 1$.
So far I have tried the following:
- First, I use the map $z\mapsto \frac 1z$, so $D$ is conformally equivalent to $D_1 = \{z\in\Bbb C : |Arg(z)|<\frac \pi4\}\setminus[0,1]$.
- Then, I use the map $z\mapsto z^4$ to 'close' the sector and have $D_2 = \Bbb C\setminus (-\infty,1]$.
- Now, I take the map $z\mapsto z-1$ to move $D_2$ into $D_3 = \Bbb C\setminus (-\infty,0]$.
- I 'reopen' $D_3$ with $z\mapsto \sqrt z$ and get $D_4 = \{z : Re(z) >0\}$.
- Finally, I use the Möbius transformation $T(z) = \frac{1+z}{1-z}$, and reach to the unit disc $\Bbb D$.
Of course, this does not verify that $f(z_0) = 0$, so I tried changing the last step in order for it to be verified, so an alternate Möbius transformation would be $T(z) = \frac{-\sqrt{15}+z}{1-z}$, but I'm sure that this does not give me the conformal map that I'm looking for either.
Could anyone please help me out? Am I complicating this too much, or can it be changed so the computation gets easier?