# How to find a conformal map from the upper half-plane to this L-shaped domain?

I want to find an integral representation of the conformal map that sends the upper half plane $\text {Im}z>0$ onto the infinite L-shaped region $$\Omega = \{z = x+iy; \ x > 0, \ y > 0, \ \min(a,\frac yb) < 1 \}$$

which is a slight modification of this question, where we change the vertex of the turn from the point $1+i$ to the more general point $1+bi$ in the first quadrant. (This is an exercise in Gamelin's text)

By the Schwarz-Christoffel formula, the solution is of the form $$F(w) = K\int_{z_1}^w (\zeta-z_0)^{-1}(\zeta-z_1)^{-1/2}(\zeta-z_2)^{-1} \ d\zeta$$ where $z_0, z_1, z_2$ are the numbers that get mapped by the conformal mapping to the vertices at $i \infty, 0, +\infty$ respectively. I do not know what are these preimages of the vertices.

In the linked question they are given as $+1, -1, 0, \infty$ for the particular case.

Ahlfors says in his book that in general there is no formula for determining the prevertices, and only $3$ of them can be arbitrarily chosen. Since we have $4$ points to determine in this case, we may assume that $3$ of the prevertices are $z_0=-1, z_1=0, z_2=\infty$.

I was told that the right decision for the remaining vertex is $b^2$. How can I get this? Is there an intuitive explanation for why this is the right choice of the prevertex $z_2$, perhaps using the symmetry of the domain along the line $y=bx$?

The L-shaped domain has four vertices $0,\infty, 1+bi$ and $i\infty.$ As Ahlfors says, we can choose $3$ of prevertices aibitrarily, so we may assume$$z=-1\longleftrightarrow w=i\infty,\quad z=0\longleftrightarrow w=0,\quad z=\infty \longleftrightarrow w=1+bi.$$ Let the remaining prevertex be $z=a$ (it's value is unknown at the present time). Then the mapping function has the form
$$F(z) = K\int_{0}^z (\zeta+1)^{-1}\zeta^{-1/2}(\zeta-a)^{-1} \ d\zeta.\tag{1}$$ We have to determine the value of $a$ so that $F(\infty)=1+bi.$ \begin{align} F(\infty)&=F(i\infty)= K\int_0^\infty \frac{idt}{(it+1)\sqrt{it}(it-a)} \\ &= -\frac{K}{\sqrt{2}}\int_0^\infty \frac{(1+i)(t^2+a+it(1-a))}{(t^2+1)\sqrt{t}(t^2+a^2)}dt \\&=I+J+i(I-J), \end{align} where $$I=-\frac{K}{\sqrt{2}}\int_0^\infty \frac{t^2+a}{(t^2+1)\sqrt{t}(t^2+a^2)}dt,\quad J=-\frac{K}{\sqrt{2}}\int_0^\infty \frac{(a-1)\sqrt{t}}{(t^2+1)(t^2+a^2)}dt.$$ Since $F(i\infty)$ should be $1+bi,$ $K,$ $a$ should be determined so that $$I+J=1,\quad I-J=b$$ holds. I could not get $a=b^2$, however $(1)$ is an integral expression for $w=F(z)$.
EDIT: $$a=\frac{1}{b^2}$$ will be true, not $a=b^2.$
In the case $b=1$, we can find $F(z)$ explicitly. When $b=1$, we have $a=1.$ Thus $$-\frac{K}{\sqrt{2}}\int_0^\infty \frac{dt}{(t^2+1)\sqrt{t}}=-\frac{\pi K}{2}=1$$ gives $K=-\frac{2}{\pi}$ and we get explicitly $$F(z)=-\frac{2}{\pi}\int_0^z \frac{d\zeta }{(\zeta^2 -1)\sqrt{\zeta }}=\frac{1}{\pi}\left(\log \frac{1+\sqrt{z}}{1-\sqrt{z}}+\tan^{-1}\sqrt{z}\right).$$
• Thank you. Did you get $a=\frac{1}{b^2}$ by directly doing the the long integral $I$? Perhaps this result for the value of $a$ can be interpreted somehow, heuristically? (There is some calculation considering the residue of the integrand at the prevertices in the linked question that I do not understand but might be relevant. Calculating $F(z)$ explicitly seems not much harder than finding the integral representation, which makes it strange if that's Gamelin's intention) – Emolga Apr 20 '17 at 23:39
• Directly doing the looong integral...,using resdue theorem. I'm now looking for easier ways to prove $$\int_0^\infty \frac{t^2+a+(a-1)t}{(t^2+1)\sqrt{t}(t^2+a^2)}dt=\sqrt{a}\int_0^\infty \frac{t^2+a-(a-1)t}{(t^2+1)\sqrt{t}(t^2+a^2)}dt.$$ – ts375_zk26 Apr 21 '17 at 2:31