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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.

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

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  • $\begingroup$ Wow! wonderful. I will study your solution step by step to understand it completely. Thank you very much. $\endgroup$ – Bumblebee Oct 23 '17 at 14:13
  • $\begingroup$ sorry, one more question: How did you obtain the equation for $\alpha$ ? $\endgroup$ – Bumblebee Oct 23 '17 at 16:14
  • $\begingroup$ 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, $$G^\prime(z)=A\frac{(z-1)^\frac{3}{4}(z+1)^\frac{3}{4}}{(z-\alpha )^2(z-\bar{\alpha })^2}.$$ $$ \operatorname{Res}(G^\prime,\alpha )=\lim_{z\to \alpha }\frac{d}{dz}\left\{(z-\alpha )^2G^\prime(z)\right\}=0\implies \frac{3}{4}\left(\frac{1}{\alpha +1}+\frac{1}{\alpha -1}\right)-\frac{2}{\alpha -\bar{\alpha }}=0.$$ $\endgroup$ – ts375_zk26 Oct 24 '17 at 0:21

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