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This comes straight from Conway's Complex Analysis, VII.4, exercise 4.

Find an analytic function $f$ which maps $G:=$ {${z: |z| < 1, Re(z) > 0}$} onto $B(0; 1)$ in a one-one fashion.

$B(0;1)$ is the open unit disc.

My first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \subset B(0;1)$. Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.

My next thought is to modify $z^2$, something like $a(z-h)^2+k$. I've yet to work out the details, but my gut tells me this isn't the right idea.

I've been teaching myself conformal maps in preparation for a qualifying exam. So, if there's a shockingly basic, obvious solution... please patronize me.

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    $\begingroup$ One thought (which I mostly pose from intuition) is seek a conformal map which maps $G$ to the right half-plane; from there it's not hard to map the right-half plane to $B(0;1)$ by a rotation. $\endgroup$ – Semiclassical Jul 30 '14 at 2:20
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The following trick works for any region bounded by two circular arcs (or a circular arc and a line).

Find the points of intersection of the arc and the line. (Here, they're $i$ and $-i$.) Now pick a Mobius transformation that takes one of those points to $0$ and the other to $\infty$; here $z \mapsto \frac{z-i}{z+i}$ works. Then the arc and the line go to two rays (because a Mobius transformation sends circles in $S^2$ to circles in $S^2$, and the only circles in $S^2$ that goes through both $0$ and $\infty$ is a line in $\Bbb C$), both starting at $0$ and going off the $\infty$. Your domain maps to the region bounded by these two rays.

Let's compute the rays. It suffices to find where a single point on each arc maps; if $z_0$ is on the arc, the ray will be $\{f(z_0)t : 0 \leq t < \infty\}$. I say we pick $0$ to be our point of choice on $Re(z) = 0$ and $1$ to be the point of choice for the circular arc. These are mapped to $-1$ and $-I$ respectively; so our two arcs are the negative real axis and the negative imaginary axis. I'd like the "lower" arc to be the positive real axis, so let's multiply by $-1$ to do this.

So we have a conformal map from your half-disc to the upper-right quadrant given by $z \mapsto -\frac{z-i}{z+i}$. The upper half-plane is nicer, so let's map to that by squaring; now we have a map to the upper half plane given by $z \mapsto \frac{(z-i)^2}{(z+i)^2}$. (For other regions bounded by rays that make different angles, you get to the upper half plane by a $z \mapsto z^\beta$ for the appropriate $\beta$.)

Now there's a standard map from the upper half plane to the unit disc given by $z \mapsto \frac{z-i}{z+i}$. Composing this with our last map gives us a map from the semi-disc to the unit disc, given by $$z \mapsto -i\frac{z^2+2z-1}{z^2-2z-1}.$$

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  • $\begingroup$ This is more insightful. I'll delete my answer. $\endgroup$ – user138530 Jul 30 '14 at 3:06
  • $\begingroup$ For fun, I added the images of your mappings as a community wiki answer. Enjoy! $\endgroup$ – Semiclassical Jul 30 '14 at 3:15
  • $\begingroup$ @ChristianRemling I appreciate the compliment! $\endgroup$ – user98602 Jul 30 '14 at 3:15
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    $\begingroup$ It might be worth noting that, even before computing the two rays (the negative real and imaginary axes), you knew that they had to be perpendicular, because conformal maps preserve angles. So you already know that the next step should be squaring and that it would produce a half-plane. The details of the rays and the half-plane are needed only at the very end, when you map the half-plane to the disk. $\endgroup$ – Andreas Blass Jul 30 '14 at 4:04
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    $\begingroup$ @AndreasBlass That's definitely worth noting - thanks for commenting. $\endgroup$ – user98602 Jul 30 '14 at 4:31
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As a supplement to the fine answers provided, here are the pictures of the conformal maps themselves:

enter image description here

We start with the right half of the unit disk, map it to the upper right quadrant, square this to the upper half-plane, and finally rotate onto the unit disk.


Per requests for code - the basic pattern to generate the image of a polar region $a\leq r \leq b$ and $\alpha \leq \theta \leq \beta$ under a function $f$ in Mathematica is:

ParametricPlot[{Re[f[r*Exp[I*theta]]], Im[f[r*Exp[I*theta]]]},
  {r, a, b}, {theta, alpha, beta}, Mesh -> True]

The pictures above may be generated with the following code block:

phi1[z_] = -(z - I)/(z + I);
f[z_] = z^2;
phi2[z_] = (z - I)/(z + I);
nestedFunctions = 
  Composition @@@ Table[Take[{phi2, f, phi1}, -k], {k, 0, 3}];
pics = Table[
  ParametricPlot[{Re[g[r*Exp[I*theta]]], Im[g[r*Exp[I*theta]]]},
      {r, 0, 1}, {theta, -Pi/2, Pi/2}, 
      Mesh -> True, PlotRange -> 2, ImageSize -> 360],
    {g, nestedFunctions}];
Grid[Partition[pics, 2]]

You can find a lot more information on visualizing complex function with Mathematica in this notebook.

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  • $\begingroup$ It appears I can only choose one answer, but I think these images are just as essential to the solution as the explanation. With what program did you generate them? $\endgroup$ – artificial_moonlet Jul 30 '14 at 15:06
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    $\begingroup$ It's from Mathematica. I'll incorporate the code for them when I get the chance. (Also, I should note that this really isn't a proper answer since I didn't remember the proper conformal maps until I saw the other answers. Hence why I did this as a community wiki answer.) $\endgroup$ – Semiclassical Jul 30 '14 at 15:08
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    $\begingroup$ As a rookie mathematica user I would find it usefully to see the code for these types of mappings. $\endgroup$ – coffeebelly Oct 4 '14 at 19:10
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    $\begingroup$ I wonder if you can still recall the code :) $\endgroup$ – snulty Jan 11 '17 at 8:15
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    $\begingroup$ @snulty Code has been posted, if you are still interested. $\endgroup$ – Mark McClure Dec 8 '18 at 17:08

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