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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.

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    $\begingroup$ My book of conformal transformations says the following does the trick: $$F(z) = 2i\frac{\tanh^{-1}\sqrt{z} - \tan^{-1}\sqrt{z}}{\pi}+1+i$$ $\endgroup$ – user98602 Jul 4 '14 at 17:09
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    $\begingroup$ Another approach cut the region along the line of symmetry, from $0$ to $1+i$. Mark the cut as a dashed line. Map halfplane onto one half; then arrange for the dashed preimage to be, say $(-\infty,0)$, and reflect across it. With one more square root you have the map. However, it does not seem to offer advantages over what you are doing, because you also exploit symmetry in your solution. And the $\pi/4$, $3\pi/4$ angles make for a difficult integral. $\endgroup$ – user147263 Jul 5 '14 at 2:00
  • $\begingroup$ In a nutshell: wrong quadrant is your fault (should have used $(1-w)$), while wrong scaling is your software's fault, not taking the right branch of logarithm. $\endgroup$ – user147263 Jul 5 '14 at 2:37
  • $\begingroup$ I don't have software, so in fact it's all my fault... ;) $\endgroup$ – bryanj Jul 5 '14 at 2:40
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    $\begingroup$ You could use Sage online... var('w') then g = (w+1)^(-1)*w^(-1/2)*(1-w)^(-1)*2/pi, then g.integral(w), and so on... Plotting will be a hassle, but there is an array of plotting functions. Just as Maple, Sage gets the branch of one of two logarithms wrong. $\endgroup$ – user147263 Jul 5 '14 at 3:00
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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));

L shaped

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.

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  • $\begingroup$ Thanks! I should've seen closed form for the integral, which you gave, and which agrees with the one in Mike Miller's comments. Substitute $u = \sqrt{w}$, then use partial fractions twice. $\endgroup$ – bryanj Jul 5 '14 at 3:30

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