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.
 A: 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 $G: H\to \mathbb{\overline C}\setminus T.$
