# From the unit disk to the right half plane and $T(0)=3$

Find a Mobius transformation $T$ from the unit disk to the right half plane with condition $T(0)=3$.

First, the transformation from the unit circle to the upper half plane is $T_1(z)=(1-i)\frac{z-i}{z-1}$.

So from the unit circle to the right half plane, $T_2(z)=-i(1-i)\frac{z-i}{z-1}$

How can I introduce the condition $T(0)=3$ ?

$T(0)=1-i\neq3$

• @WillJagy how do you introduce the condition $T(0)=3$?
– user486983
Sep 13 '18 at 21:45
• It seems that you already anticipate that Möbius transformations will play a role in the solution of this problem. Connected with the idea that these "linear fractional transformations" map lines and/or circles to lines and/or circles is a way to characterize them by the images of three non-colinear points in the extended complex plane, usually $0,1,$ and $\infty$. Sep 16 '18 at 18:04

For a purely algebraic derivation, consider the general form of the Möbius transformation $$\,T(z)=\dfrac{az+b}{cz+d}\,$$. Both $$\,a\,$$ and $$\,c\,$$ cannot be $$\,0\,$$, otherwise it would be a constant transformation. The right half-plane is invariant to the inversion $$\,T(z) \to \dfrac{1}{T(z)}\,$$ so it can be assumed WLOG that $$\,a \ne 0\,$$, then after normalizing it can be assumed WLOG that $$\,a=1\,$$. The condition $$\,T(0)=3\,$$ translates to $$\,b = 3d\,$$, so in the end $$\,T(z)=\dfrac{z+3d}{cz+d}\,$$ for some $$\,c, d \in \Bbb C\,$$ with $$\,d \ne 0\,$$.

The unit circle must transform into the imaginary axis, so for $$\,|z|=1\,$$:

\begin{align} 0 = 2 \operatorname{Re}\left(T(z)\right) &= \dfrac{z+3d}{cz+d} + \dfrac{\bar z+3 \bar d}{\bar c \bar z+ \bar d} \\ &= \frac{(z+3d)(\bar c \bar z + \bar d)+(\bar z + 3 \bar d)(cz + d)}{|cz+d|^2} \\ &= \frac{(c+\bar c)|z|^2+ 6 |d|^2+(\bar d + 3c\bar d) z+(3 \bar cd +d)\bar z}{|cz+d|^2} \\ &= \frac{2 \operatorname{Re}(c)+ 6 |d|^2+(3c+1)\bar d z+(3 \bar c +1)d\bar z}{|cz+d|^2} \end{align}

It follows that $$\,3c+1=0\,$$ for the numerator to not depend on $$\,z\,$$, and $$\,2 \operatorname{Re}(c)+ 6 |d|^2=0\,$$ for the numerator to be $$\,0\,$$. The first equation gives $$\,c = -\dfrac{1}{3}\,$$, and the second one $$\,|d|=\dfrac{1}{3}\,$$. Therefore, defining $$\,\omega = 3d\,$$ the general solution is:

$$T(z) \;=\; \frac{z + 3d}{-\dfrac{1}{3}z+d} \;=\; 3\,\dfrac{z + \omega}{-z + \omega} \quad\quad\style{font-family:inherit}{\text{where}}\;\; |\omega|=1$$

[ EDIT ]   For quick verification of the form above:

$$\small \frac{1}{3}T(z) = \dfrac{z + \omega}{-z + \omega} \color{red}{\cdot \frac{\bar \omega}{\bar \omega}} = \frac{1+\bar \omega z}{1 - \bar \omega z} \color{red}{\cdot \frac{1 - \omega \bar z}{1 - \omega \bar z}} = \frac{1 - |\omega|^2|z|^2+\bar \omega z - \omega \bar z }{|1 - \bar \omega z|^2} = \frac{1 - |z|^2+ 2i \operatorname{Im}(\bar \omega z)}{|1 - \bar \omega z|^2}$$

Therefore $$\,\small\operatorname{Re}(T(z)) = 3\,\dfrac{1 - |z|^2}{|1 - \bar \omega z|^2} \ge 0\,$$ iff $$\,\small|z| \le 1\,$$, and of course $$\,\small T(0) = 3\,$$.

• Why the right half-plane is invariant to the inversion $T(z)\to\frac{1}{T(z)}$ what do you mean?
– user486983
Sep 22 '18 at 5:14
• @Isabella Context was that (at least) one of $\,a,c\,$ must be non-zero. But $\,w = T(z)\,$ is in the right half-plane iff $\,\frac{1}{w} = \frac{1}{T(z)}\,$ is in the right half-plane, since $\,\operatorname{Re}\left(\frac{1}{x + i y}\right) = \frac{x}{x^2+y^2}\,$, and so it is WLOG to assume that the non-zero coefficient is the one in the numerator, otherwise the same argument would apply to $\,\frac{1}{T(z)}\,$. Of course, it is easy to show that both $\,a,c\,$ must be non-zero, but that would have been more than technically needed here.
– dxiv
Sep 22 '18 at 5:27
• I don't understand, I'll read it again later
– user486983
Sep 22 '18 at 5:51
• @Isabella If $\,T(z)=\frac{az+b}{cz+d}\,$ maps a point $\,z\,$ to the right half-plane, then so will $\,U(z)=\frac{1}{T(z)}=\frac{cz+d}{az+b}\,$, since the real parts $\,\operatorname{Re}(U(z)) = \operatorname{Re}\left(\frac{1}{T(z)}\right)=\frac{\operatorname{Re}(T(z))}{|z|^2}\,$ have the same sign. So, if we know that (at least) one of $\,a,c\,$ is non-zero, there is no loss of generality to assume that the one in the numerator $\,a \ne 0\,$, since the same argument would work for the transformation $\,U(z)=\frac{1}{T(z)}\,$ otherwise.
– dxiv
Sep 22 '18 at 6:06

Let $$\mathbb D$$ be the open unit disc, $$U$$ the open right half plane. Suppose $$f:\mathbb D\to U$$ is bilholomorphic, with $$f(0)= 1-i$$ (just as your map $$T_2$$ does.) How do we then find a biholomorphic map $$g:\mathbb D\to U$$ with $$g(0)= 3?$$

There are two families of biholomorphic maps from $$U$$ to $$U$$ that are simple and will be helpful here:

i) Vertical translations. These are the maps $$v_c(z)=z+ic,$$ where $$c$$ is a real constant.

ii) Positive dilations: These are the maps $$d_r(z)= rz,$$ where $$r$$ is a positive real constant.

Claim: $$g(z) = (d_3\circ v_{1}\circ f)(z)$$ has the desired properties.

Proof: It should be clear that $$g:\mathbb D\to U$$ is biholomorphic. And we see

$$g(0) = d_3(v_1(f(0)))) = d_3(v_1(1-i)) = d_3((1-i)+i))= d_3(1)=3.$$

So $$g$$ does the job. For completeness, note $$g(z) = 3(f(z) +i).$$

• I simplified my answer and I think improved it.
– zhw.
Sep 21 '18 at 23:59
– user486983
Sep 23 '18 at 16:39
• I showed that if $f:\mathbb D\to U$ is biholomorphic, and $f(0)=1-i,$ then $g(z)=3(f(z)+i)$ is a biholomorphic map from $\mathbb D$ to $U$ that sends $0$ to $3.$ Since you already found such an $f$ (your map $T_2$), I thought that using your $f$ would be a simple way to go. Continued below ...
– zhw.
Sep 23 '18 at 18:50
• But note that a map with these properties is not unique. In fact, if $g$ is one such map, then so is $g(cz),$ where $c$ is a constant with $|c|=1.$ Furthermore, all such maps are of the form $g(cz),|c|=1.$ For example, one of the other anwers has $g(z) = 3(1+z)/(1-z)$ as such a map. You can verify that the map I gave in my answer is $g(-iz)$ for this $g.$
– zhw.
Sep 23 '18 at 18:50
• me? verify that the map you gave in your answer is $g(iz)$ for $g$? I am not the one receiving 100 bounty..
– user486983
Sep 24 '18 at 15:22

You took only one possible transformation that maps the unit disk to the right half-plane. Generally, since the conjugation wrt the unit circle becomes the conjugation wrt the imaginary axis, $$T(0) = 3$$ implies $$T(\infty) = -3$$. Therefore, $$T(z) = 3 - 6 z/(z + a)$$. Then $$T(a) = 0$$, therefore $$|a| = 1$$ gives the general form of $$T$$.

It is well worth knowing that the conformal map $z \rightarrow \frac{1+z}{1-z}$ permutes the regions shown below $(1234)$ and act similarly on the lower half plane.

In particular the unit circle is mapped to the right half plane, indeed one can check that the unit circle ($z=e^{i \theta} =\cos( \theta)+i \sin( \theta)$ is mapped to the imaginary axis \begin{eqnarray*} \frac{1+z}{1-z}= \frac{1+\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)-i \sin( \theta)} \times \frac{1-\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)+i \sin( \theta)} = \cdots \end{eqnarray*} Now in order to fulfill the requirement that $T(0)=3$ simply multiply this function by $3$ (and note that this will still map the unit circle to the imaginary axis) & so ...

\begin{eqnarray*} T(z)=3 \frac{1+z}{1-z} \end{eqnarray*} will do the trick.

• How did you find it?
– user486983
Sep 13 '18 at 21:59
• We want a conformal map that will map $e^{i \theta}$ to a purely imaginary value ... use DeMoivre's theorem to check this, & then just rescale the map by a factor of $3$. Sep 13 '18 at 22:03
• Is DeMoivre's theorem involved here? really?
– user486983
Sep 13 '18 at 22:08
• I mean $z=e^{i \theta} =\cos( \theta)+i \sin( \theta)$ ... now multiply top & bottom by the conjuagate of the denominator. Sep 13 '18 at 22:11
• hmm is it too much to ask you to write a detailed answer? :)
– user486983
Sep 13 '18 at 23:18