Within $\mathbb{C} \cup \{ \infty \}$, consider the unit-disk $\mathbb{D} = \{ z : |z|\leq 1 \}$ with three points labelled as $a$, $b$, $c$ on its boundary.

I want to map $\mathbb{D}$ conformally and bijectively to the upper-half plane $\mathbb{H} = \{ z : Im(z) \geq 0\}$ with the added constraint that $a$, $b$ and $c$ should be mapped to $0$, $1$ and $\infty$ respectively.

Is this possible?

With the moebius map $f(z) = \frac{i (1+ z)}{1-z}$ we can map $\mathbb{D}$ to $\mathbb{H}$.
But I am not sure, how to enforce the mapping of the points $a$, $b$ and $c$ to $0$, $1$ and $\infty$ respectively.

Do we compose this $f$ with another moebius map?


$S(z)=\frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$ is the mobius transformation that maps $z_2\to 1$, $z_3\to 0$ and $z_4\to \infty$.

By orientation principle, a Möbius Transformation(such as $S(z)$) with $S(\Gamma_1)=\Gamma_2$maps the left (or right) hand side of $\Gamma_1$(with respect to ($z_1,z_2,z_3)$) to left (or right) hand side of $\Gamma_2$(with respect to the orientation $(Sz_1,Sz_2,Sz_3)$).

enter image description here

Basically what you want to do is find a map, $T$, which takes $1\to -1$, $i\to 0$ and $-1\to 1$

You can now construct a Möbius Transform $M$ that would take $1\to 1$, $i\to 0$ and $-1\to \infty$

Also, $N$ which would take $-1\to 1$, $0\to 0$ and $1\to \infty$

Notice: Will $T(z)=N^{-1}\circ M(z)$ do the trick?

  • $\begingroup$ There has been a little error in the right figure. It should read $-1,0,1$ and not $-i,0,i$ $\endgroup$ – Swapnil Tripathi Nov 20 '14 at 19:40

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