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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 $T(0)=3$ ?

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

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  • $\begingroup$ @WillJagy how do you introduce the condition $T(0)=3$? $\endgroup$ – user486983 Sep 13 '18 at 21:45
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    $\begingroup$ 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 $\infty$. $\endgroup$ – hardmath Sep 16 '18 at 18:04
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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 $\,\small T(0) = 3\,$.

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  • $\begingroup$ Why the right half-plane is invariant to the inversion $T(z)\to\frac{1}{T(z)}$ what do you mean? $\endgroup$ – user486983 Sep 22 '18 at 5:14
  • $\begingroup$ @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. $\endgroup$ – dxiv Sep 22 '18 at 5:27
  • $\begingroup$ I don't understand, I'll read it again later $\endgroup$ – user486983 Sep 22 '18 at 5:51
  • $\begingroup$ @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. $\endgroup$ – dxiv Sep 22 '18 at 6:06
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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 $g(z) = 3(f(z) +i).$

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  • $\begingroup$ I simplified my answer and I think improved it. $\endgroup$ – zhw. Sep 21 '18 at 23:59
  • $\begingroup$ Why your final answer does not match with the other given answers? $\endgroup$ – user486983 Sep 23 '18 at 16:39
  • $\begingroup$ 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 ... $\endgroup$ – zhw. Sep 23 '18 at 18:50
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    $\begingroup$ 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 $g.$ $\endgroup$ – zhw. Sep 23 '18 at 18:50
  • $\begingroup$ me? verify that the map you gave in your answer is $g(iz)$ for $g$? I am not the one receiving 100 bounty.. $\endgroup$ – user486983 Sep 24 '18 at 15:22
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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 $T$.

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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.

enter image description here

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.

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  • $\begingroup$ How did you find it? $\endgroup$ – user486983 Sep 13 '18 at 21:59
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    $\begingroup$ 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 $3$. $\endgroup$ – Donald Splutterwit Sep 13 '18 at 22:03
  • $\begingroup$ Is DeMoivre's theorem involved here? really? $\endgroup$ – user486983 Sep 13 '18 at 22:08
  • $\begingroup$ I mean $z=e^{i \theta} =\cos( \theta)+i \sin( \theta)$ ... now multiply top & bottom by the conjuagate of the denominator. $\endgroup$ – Donald Splutterwit Sep 13 '18 at 22:11
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    $\begingroup$ hmm is it too much to ask you to write a detailed answer? :) $\endgroup$ – user486983 Sep 13 '18 at 23:18

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