Schwarz-Christoffel mapping onto infinite L-shaped region

Question

I'm trying to map the upper half plane onto the infinite L-shaped region $$ \Omega = \{z = x+iy; \ x > 0, \ y > 0, \ \min(x,y) < 1 \} $$ My first try is a Schwarz-Christoffel function $$ F(w) = \int_0^w (\zeta+1)^{-1}\zeta^{-1/2}(\zeta-1)^{-1} \ d\zeta $$ This guy looks promising, because the integrand seems to have the right turning angles at $\zeta = -1, 0, \text{ and}\ 1$; in addition $F(w)$ seems to satisfy the symmetry $$ F(-x + iy) = i \overline{F(x + iy)} $$ Unfortunately, the image appears to be scaled wrong (and lives in the third quadrant).

Question: How can I find the negative real constant by which to flip & scale it to the correct position?

Guess: Calculate $F(\infty)$ by integrating along the positive imaginary axis, then my multiplier will be $\displaystyle \frac{-\sqrt{2}}{|F(\infty)|}$.

Thanks for any tips, especially including a better way to approach the problem.

Answer 1

So, you want to send $-1,0,1$ into $i\infty, 0, +\infty$, thus forming the outer boundary of the $L$-shape. The inner boundary will be formed by $(1,\infty)$ going to $(+\infty+i,1+i)$, and by $(-\infty, -1)$ going to $(1+i, 1+i\infty)$. Makes sense enough. Then the integral should be
$$F(w) = K\int_0^w (\zeta+1)^{-1}\zeta^{-1/2}(1-\zeta)^{-1} \ d\zeta \tag1$$ because we want $F'>0$ when $0<w<1$. Next, $K$ should be $2/\pi$ because when crossing $w=1$ we pick up $-\pi i$ times the residue there (upper semicircle traced clockwise), and this amount is $$-\pi i\, \operatorname{res}_{\zeta=1}(\zeta+1)^{-1}\zeta^{-1/2}(1-\zeta)^{-1} = \frac{\pi i}{2}$$ Something like Maple will give you an antiderivative of (1) at once: $$ F(w) = \frac{1}{\pi} \log(1+\sqrt{w}) - \frac{1}{\pi} \log(1-\sqrt{w}) + \frac{2}{\pi} \arctan(\sqrt{w}) $$ but will probably not place branch cuts where you want them. To help the stupid program, I replaced $w$ with $\sqrt{r}e^{it/2}$ and then plot using a value of $t$ slighly off Maple's branch cut line.

g:=(1/Pi)*ln(1+sqrt(r)*exp(I*t/2)) - (1/Pi)*ln(1-sqrt(r)*exp(I*t/2)) + (2/Pi)*arctan(sqrt(r)*exp(I*t/2));
display(complexplot(eval(g,t=0.00001),r=0..5,thickness=3), complexplot(eval(g,t=Pi),r=0..5,thickness=3));

Never mind that the figure appears to be cut off on top and to the right; computer has a hard time grasping the unboundedness of logarithm.