# Conformal mapping from upper half plane to a triangle

Let $T$ be the closed triangle with vertices at complex numbers $0, i, \dfrac12+\dfrac{i}2.$ How do we find explicit formula for the conformal mapping from upper half plane to $\Bbb{\overline C}\setminus T$?

I know this is an application of Schwarz–Christoffel mappings, but do not know how to it works with these complex numbers.

I. Explicit formula for the conformal mapping from the upper half plane $H$ to $T$:
$T$ has three vertices $w=i,\,0,\, 1/2+i/2$ with angles $\pi/4,\,\pi/4,\,\pi/2$ respectively. We assume $$w=i\longleftrightarrow z=-1,\quad w=0\longleftrightarrow z=1,\quad w=\frac{1}{2}+\frac{i}{2}\longleftrightarrow z=\infty.$$ By the Schwarz-Christoffel formula the mapping function $w=F(z)$ will be of the form $$F(z)=i+A\int_{-1}^z(\zeta -1)^{\frac{1}{4}-1}(\zeta +1)^{\frac{1}{4}-1}d\zeta .$$ The constant $A$ should be determined so that $F(1)=0.$
II. Explicit formula for the conformal mapping from $H$ to $\mathbb{\overline C}\setminus T$:
This time we assume $$w=0\longleftrightarrow z=-1,\quad w=i\longleftrightarrow z=1,\quad w=\frac{1}{2}+\frac{i}{2}\longleftrightarrow z=\infty.$$ Then by the Schwarz-Christoffel formula the mapping function $w=G(z):H\to \mathbb{\overline C}\setminus T$ will be of the form $$G(z)=A\int_{-1}^z \frac{(\zeta -1)^{\color{red} {1-\frac{1}{4}}}(\zeta +1)^{ \color{red}{1-\frac{1}{4}}}}{\color{red}{(\zeta -\alpha )^2(\zeta -\bar{\alpha })^2}}d\zeta ,$$ where $z=\alpha$ is a point at which $G(z)$ has a simple pole ($G(\alpha )=\infty$). Since $G(z)$ has a simple pole at $z=\alpha$, $\alpha$ has to satisfy $$\frac{3}{4}\left(\frac{1}{\alpha +1}+\frac{1}{\alpha -1}\right)-\frac{2}{\alpha -\bar{\alpha }}=0.$$ Thus we get $\alpha =i\sqrt{2}$ and $G(z)$ will be $$G(z)=A\int_{-1}^z \frac{(\zeta -1)^\frac{3}{4}(\zeta +1)^\frac{3}{4}}{(\zeta ^2+2)^2}d\zeta .$$ The constant $A$ should be determined so that $G(1)=i.$ After some calculations we have $$G(z)=\frac{8{\mit\Gamma}(\frac{5}{4})e^{-\frac{\pi i}{4}}}{\sqrt{\pi}{\mit\Gamma}(\frac{3}{4})}\int_{-1}^z \frac{(\zeta -1)^\frac{3}{4}(\zeta +1)^\frac{3}{4}}{(\zeta ^2+2)^2}d\zeta .$$ This is the explicit formula for the conformal mapping $G: H\to \mathbb{\overline C}\setminus T.$
• sorry, one more question: How did you obtain the equation for $\alpha$ ? – Bumblebee Oct 23 '17 at 16:14
• the equation for $\alpha$. Since $G(z)$ has a simple pole at $z=\alpha$, $G(z)=\frac{c_{-1}}{z-\alpha }+c_0+c_1(z-\alpha )+\cdots,$ $\operatorname{Res}(G^\prime,\alpha )=0$. On the contrary, $$G^\prime(z)=A\frac{(z-1)^\frac{3}{4}(z+1)^\frac{3}{4}}{(z-\alpha )^2(z-\bar{\alpha })^2}.$$ $$\operatorname{Res}(G^\prime,\alpha )=\lim_{z\to \alpha }\frac{d}{dz}\left\{(z-\alpha )^2G^\prime(z)\right\}=0\implies \frac{3}{4}\left(\frac{1}{\alpha +1}+\frac{1}{\alpha -1}\right)-\frac{2}{\alpha -\bar{\alpha }}=0.$$ – ts375_zk26 Oct 24 '17 at 0:21