Show that $u(r,\theta) = \frac{1}{\pi}\arctan\left(\frac{1-x^2-y^2}{(x-1)^2+(y-1)^2-1}\right)\\$ where $\arctan(t) \in [0,\pi]$ is the solution to Dirichlet's problem for a unit disk for the piecewise continuous function: $h(e^{i\phi}) = \begin{cases} 0&\text{if}\, \phi \in (0, \pi/2)\\ 1&\text{if}\, \phi \in (\pi/2, 2\pi)\\ \end{cases}$
(b) Using a biholomorphic map to change the region to the upper half plane
So for some context: part (a) of this problem asked to show the same thing using the Poisson kernel. I did this by defining the function $u(r,\theta) = \frac{1}{2\pi}\int_0^{2\pi} P_r(\phi-\theta) h(e^{i\phi})\,d\phi = \frac{1}{2\pi}\int_{\pi/2}^{2\pi} P_r(\phi-\theta) \,d\phi$, then integrating $P_r(\psi)$ using the formula: $\frac{d}{d\psi}\left(2\arctan\left(\frac{1+r}{1-r}\tan{\frac{\psi}{2}}\right)\right) = \frac{1-r^2}{1-2r\cos(\psi)+r^2} = P_r(\psi)$ and arriving at the correct $u(r,\theta)$.
So for part (b) so far I have used the Mobius transformation $M(z)=\frac{i(z+1)}{1-z}$ as a biholomorphic map from the disk to the upper half plane. Then $M(h(e^{i\phi}))= \begin{cases} M(0)=i&\text{if}\, \phi \in (0, \pi/2)\\ M(1)=\infty&\text{if}\, \phi \in (\pi/2, 2\pi)\\ \end{cases}$ but now I am not sure how to proceed. How do I "redo" part (a) with the changed region? Do I reevaluate the integral using the biholomorphic map M(z) instead of $h(e^{i\phi})$ or am I not understanding the problem correctly?
We use the biholomorphic transformation: $\beta: \Delta \rightarrow \mathbb H$ given by $\beta(z) = (1 - i)\frac{z-i}{z+i}$ which sends 1 to $\infty$, i to 0, and -1 to 1.
Transforming the answer from part (a) will help us work backwards: $u(r,\theta) = \frac{1}{\pi}\arctan\left(\frac{1-x^2-y^2}{(x-1)^2+(y-1)^2-1}\right)\\$ is the same as $\frac{1}{\pi}\arctan\left(\frac{v}{u}\right)\\$ where $w = u + iv$.
Then we transform $\beta$ into w: $w = (1 - i)\frac{z-i}{z-1} = (1 - i)\frac{(z-i)(\overline{z}-1)}{|z-1|^2} = (1 - i)\frac{i+|z|^2-z-i\overline z}{|z-1|^2} = \frac{1+i+|z|^2-i|z|^2-z+iz-\overline z-i\overline z}{|z-1|^2} = \frac{1+i+|z|^2-i|z|^2-2Re(z)-2Im(z)}{|z-1|^2} = \frac{1+x^2+y^2-2x-2y+i(1-x^2-y^2)}{2x^2+2(y-1)^2} = \frac{(x-1)^2+(y-1)^2-1+i(1-x^2-y^2)}{2x^2+2(y-1)^2} = \frac{(x-1)^2+(y-1)^2-1}{2x^2+2(y-1)^2} + i\frac{(1-x^2-y^2)}{2x^2+2(y-1)^2} = u + iv$.
Therefore: $u = \frac{(x-1)^2+(y-1)^2-1}{2x^2+2(y-1)^2}$ and $v = \frac{(1-x^2-y^2)}{2x^2+2(y-1)^2}$, so the solution is:
$\frac{1}{\pi}\arctan\left(\frac{v}{u}\right)\\$ = $\frac{1}{\pi}\arctan\left(\frac{1-x^2-y^2}{(x-1)^2+(y-1)^2-1}\right)\\$