Find a conformal mapping $\{z \in \mathbb C : -pi/4 < \arg(z) < \pi/4 \}$ to $\mathbb{D}$ using only moebius maps.

analysiscomplex-analysissolution-verification

I have a quick question about an old post given here:

Given the sector $\{z \in \mathbb C : -\pi/4 < \arg(z) < \pi/4\}$, one wants to find a conformal map $\phi$ that maps the given sector to the unit disk. I understand the solution given in the post above, however I was curious if it is possible to find a solution only using moebius transformations? My plan was as follows:

Find a composition of conformal maps that map the boundary of the sector given by $\{z \in \mathbb C : arg(z) \in \{-\pi/4 , \pi/4 \} \}$ to the boundary of the unit disk. One could go as follows:

  1. Find $\phi_1$ that sends $(1-i,0,1+i) \mapsto (0,1,\infty)$ that maps the boundary to the imaginary axis. My calculation gave me the following
    $$
    \phi_1(z) = \frac{z-(1-i)}{z-(1+i)}\frac{1+i}{1-i}.
    $$
  2. As a second map we need to find $\phi_2$ that sends $(0,1,\infty) \mapsto (1,-i,i)$, here one can take
    $$
    \phi_2(z) = \frac{(i+1)-iz}{(i+1)-z}.
    $$
    Now concatenating both functions gives as $\phi_2 \circ \phi_1$ yields
    $$
    \phi_2 \circ \phi_1(z) = \frac{(2+i)z-(1+i)}{z+(1-i)},
    $$
    which indeed maps $(1-i,0,1+i) \mapsto (1,-i,i)$ as one can check manually. However, this seems not to be the desired map since the real line should be mapped into the unit circle. However take for example $z=5$ then wolfram alpha gives that
    $$
    \phi_2 \circ \phi_1(5) = 50/37+ 33i/37,
    $$
    which clearly lies outside of $\mathbb D$. What went wrong?

Best Answer

There is no Möbius transformation $T$ which maps that sector conformally to the unit disk. $T$ would map the boundary lines $\arg z = \pm \pi/4$ of the source domain (which meet at an angle of $\pi/2$) to arcs on the unit circle (which meet at an angle of $\pi$). But as conformal maps of the (extended) complex plane, Möbius transformation preserve angles.

In the referenced answer, $z \mapsto z^2$ doubles the boundary angle at $z=0$, and then it can be mapped to the unit disk with a Möbius transformation.

Your construction fails already in step 1. if $\phi_1$ sends $(1-i,0,1+i) \mapsto (0,1,\infty)$ then the circle through the points $(1-i,0,1+i)$ is mapped to the real axis, not the boundary of the sector.

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