I am trying to find a conformal map from $G =\{re^{i\theta}| 0<r<1, \frac{-\pi}{2} < \theta < \pi\}$ onto the unit disk.
I have an attempt but I am not sure if it is correct. Let $f_1(z) = z^2$. Then $f_1$ maps $G$ onto $\mathbb{D} \setminus [0,\infty)$ (Is this correct??)
If that is fine, then let $f_2(z) = \sqrt{z}$ where we define the log for $z \in \mathbb{C}$ such that $0 < $arg$z <2\pi$. Now let $f_3(z) = \frac{z-1}{z+1}$ which maps $\mathbb{D}\setminus [0,\infty)$ to the right half-plane. Then we rotate by $f_4(z) = iz$ and finally use $f_5(z) = \frac{z-i}{z+i}$ to map the upper half-plane to the unit disk. Composing these, we get the conformal map we seek.
If this is not correct, then any help to point me in the correct direction would be greatly appreciated.
Best Answer
Well, I don't think $w_1(z)=z^2$ is correct. Instead, try $w_1=e^{I\pi/2}z$ (rotation by $\pi/2$). Next, $w_2=w_1^{2/3}$. This will turn your region in to the upper half-ubit-disk. The, try Zhukovsky's mapping $w_3=\frac{1}{2}\Big(w_2+\frac{1}{w_2}\Big).$ This will map the upper half disk into the lower semiplane. And finally, $w_4=\frac{z+i}{z-i}$.