[Math] Find a conformal map from semi-disc onto unit disc

complex-analysisconformal-geometry

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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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 $$z \mapsto -i\frac{z^2+2z-1}{z^2-2z-1}.$$

Related Solutions

Complex Analysis – Conformal Map from the Punctured Unit Disc onto the Unit Disc

There is a surjective conformal map, though (in opposition to bijective, which is what Soarer has in mind in his answer). Just compose an holomorphic bijection from the disc to the upper half-plane with $z\mapsto e^{i z}$.

(I made a video showing the images under the map $z\mapsto\exp\frac{z-1}{z+1}$, which is a surjection from the unit disk to the punctured unit disk, of the circles centered at $0$. You can see in it how it manages to avoid the origin.) (The file will not last forever in that location... if someone can upload it to some site or another, it would be great!)

Later In any case, this is the Mathematica code I used:

Animate[
 ParametricPlot[
   With[{z = r Exp[I theta]},
   Through[{Re, Im}[Exp[(z - 1)/(z + 1)]]]
   ],
  {theta, 0, 2 \[Pi]},
  ImageSize -> Medium, PlotRange -> {{-1, 1}, {-1, 1}},
  PlotPoints -> 1000, Ticks -> False
  ],
 {r, 0, 0.999, 0.001}
 ]

punctured disk map

Complex Analysis – Conformal Map from Vertical Strip to Unit Disc

First apply $z \mapsto iz$ to map the strip onto the corresponding horizontal strip $-1 < \operatorname{Im} z < 1$.

Next step, apply $z \mapsto \exp(\frac{\pi z}2)$. This will give you the right half-plane $\operatorname{Re} z > 0$.

To finish off, the Möbius transformation $z \mapsto \dfrac{z-1}{z+1}$ takes the half-plane onto the unit disc.

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Related Solutions

Complex Analysis – Conformal Map from the Punctured Unit Disc onto the Unit Disc

Complex Analysis – Conformal Map from Vertical Strip to Unit Disc

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