Consider the unit disk $D = \{ z \in \mathbb C : |z| < 1 \}$ and the upper half disk $S = \{ z \in D : Im(z) > 0 \}$.

Then our assignment is to find a biholomorphic map $S \to D$.

Of course $z\to z^2$ won't quite work as $[0,\infty)$ isn't contained in the image. Could you give some hints please? A mapping from $D$ to the half-plane $\mathbb H$ would suffice, as I know the Möbius transformations for $\mathbb H \to D$. A plain Möbius transformation wouldn't work either though, as its inverse would take $D$ to a full disk/plane instead of the semidisk.


  • $\begingroup$ Do you have to find it explicitly? Or just prove that it exists, in that case it is just Riemann mapping theorem. $\endgroup$ – tom Apr 13 '13 at 12:05
  • $\begingroup$ Send one corner of $S$ to infinity, the other to zero using a Möbius transformation. Then $z \mapsto z^2$. $\endgroup$ – Martin Apr 13 '13 at 12:07

In this proof I'll show you how to find a conformal map from S to the upper half plane. Then you can take the inverse and compose it with a Möbius transform.

Consider the function $g(z)=\dfrac{1}{z+1}$, which is a Möbius transform, therefore it maps circles/lines to circles/lines. Since it has a pole in -1, we conclude that it maps the interval $[-1,1]$ , as well as the upper circle ( it is the set $\partial S\setminus (-1,1)$) to two lines, which intersect orthogonally at the point $g(1)=\frac{1}{2}$.

By taking the image of different points of S you can see that the image of S via g is the 4th quadrant, translated by 1/2 to the right, i.e. $\{(x,y)\in\mathbb R^2:x>\frac{1}{2},y<0\}$.

As a result, $g(z)-\frac{1}{2}$ maps S to the 4th quadrant.

Continuing, we rotate by $\pi/2$, so $e^{i\frac{\pi}{2}}(g(z)-\frac{1}{2})=i(g(z)-\frac{1}{2})$ maps S to the 1st quadrant.

Finally, by squaring, we can map the 1st quadrant to the upper half plane, in other words: $$(i(g(z)-\frac{1}{2}))^2=\dfrac{-(z-1)^2}{4(z+1)^2}$$ maps S to the upper half plane.

Note: now compose with $\dfrac{z-i}{z+i}$ which maps the upper half plane to the unit disk , to get a conformal map from S to D. What you need is the inverse of this map.

  • $\begingroup$ How did you find the map $g(z) = \frac{1}{z+1}$ ? $\endgroup$ – WLOG Jan 16 '14 at 13:16
  • 2
    $\begingroup$ You just want to send one of the points $-1$, $1$ to $\infty$. You could also work with $g(z)=1/(z-1)$. $\endgroup$ – Dimitris Jan 17 '14 at 2:54

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