Let $f$ be a holomorphic mapping from {$z:\Re(z)>0$} into itself. Let $1$ be a fixed point of $f$.

In addition suppose that $\left|\frac{f(2)-1}{f(2)+1}\right|=\frac13$. I want to show that $f(z)=\frac{az+b}{bz+a}$ where $a=1+e^{i\theta}$ and $b=1-e^{i\theta}$ for some $\theta$.

I tried looking at this problem in different ways, I just don't know what to do.

Of course if $f(z)=\frac{az+b}{cz+d}$ then I have the obvious $a+b=c+d$.

And when I look at $\left|\frac{f(2)-1}{f(2)+1}\right|=\frac13,$ this actually looks like $\left|\frac{f(2)-f(1)}{f(2)+f(1)}\right|=\frac13,$

but I don't know what to do with this last expression or if it is useful at all to write things this way and compute this expression. In general I know that a Möbius transformation has at most 2 fixed points and can be written as a composition of inversions, rotations, dilations and translations. Thanks.

  • $\begingroup$ Denoting the fixed point by "1" makes the thing reaaaally confusing. Denote it by like, $\alpha$, $\beta$, $\mu$, $\eta$, $\zeta$ or something. Whatever you want, but not $1$. =) Or are you supposing that the complex number $1$ IS a fixed point of $f$? The phrasing "Let $1$ be a fixed point of $f$" makes it look like $f$ has a fixed point named $1$. $\endgroup$ – Patrick Da Silva Jul 9 '11 at 22:20
  • $\begingroup$ @Patrick: $1$ is $1 = 1 + 0i \quad$ :) $\endgroup$ – t.b. Jul 9 '11 at 22:21
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    $\begingroup$ @Patrick: look at the desired conclusion: if $a$ and $b$ have the desired properties then $1$ is a fixed point. You're making me laugh (no offense) :) $\endgroup$ – t.b. Jul 9 '11 at 22:25
  • $\begingroup$ I know, I was just commenting the phrasing, I know it's actually $1$. Like, just saying "This phrasing is funny." Heh, well I also laughed at the sentence when I read it so let's say we're even. :P $\endgroup$ – Patrick Da Silva Jul 9 '11 at 22:28
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    $\begingroup$ Hint: Suppose you have a holomorphic function from the unit disc to itself with $f(0) = 0$. Then if $|f(z_0)| = |z_0|$ for some $z_0$, by the maximum modulus theorem $f(z) = e^{i\theta}z$ for some $\theta$. Use a properly defined Mobius transformation to reduce the problem to this situation. $\endgroup$ – Zarrax Jul 10 '11 at 0:24

First of all, I can't work in the right half plane, so let me translate the problem to the unit disk using the transformation $\phi(z) = \frac{z-1}{z+1}$ with inverse $\phi^{-1}(w) = \frac{1+w}{1-w}$. Note that $\phi(1) = 0$, $\phi(2) = \frac{1}{3}$ and $\phi(\frac{1}{2}) = - \frac{1}{3}$. In particular, the real line is mapped to itself by $\phi$. Note that $g = \phi \circ f \circ \phi^{-1}$ is a holomorphic map from the unit disk to itself and fixing the origin, so the problem begs for an application of Schwarz's lemma to $g$.

[Edit: if you don't like geometric arguments skip to the next paragraph.] Observe that the condition $\frac{|f(2)-1|}{|f(2)+1|} = \frac{1}{3}$ means that $f(2)$ lies on a circle intersecting the real line orthogonally (it is the circle of Apollonius with foci $\pm1$ and ratio of radii $1/3$) — in other words, $f(2)$ must lie on the circle $C$ passing through $2$ and $1/2$ as an easy computation shows. Now $\phi(C)$ is the circle around the origin with radius $1/3$ (since $\phi$ sends $C$ to a circle intersecting the real line orthogonally, $\phi(2) = 1/3$ and $\phi(1/2) = -1/3$).

Therefore [or, as Sam pointed out in a comment below, by checking directly that $|g(1/3)| = \frac{|f(2)-1|}{|f(2) + 1|} = \frac{1}{3}$] we have that $g:\mathbb{D} \to \mathbb{D}$ is a map fixing the origin and sending $1/3$ to some point $g(1/3) = \phi (f(2)) = \frac{e^{i\theta}}{3} \in \phi(C)$. Now Schwarz's lemma tells us that $g(z) = e^{i\theta}z$, and now $f(z) = \phi^{-1}\circ g \circ \phi (z)$ has the desired form as you can compute immediately.

  • $\begingroup$ Beautiful answer! So the trick is to always try to reduce to something simple where we can work on, like the unit disk? $\endgroup$ – user786 Jul 10 '11 at 1:26
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    $\begingroup$ @user786: Thank you! Well, it depends: I admit that I was stumped for a while because I misread the question... I'm comfortable with the upper half plane and the unit disk, but thinking about the condition given ($f(2) \in C$, it seemed best to go to the unit disk and trying to apply Schwarz's lemma. Also the idea of the circle of Apollonius is something to keep in mind. One can avoid many cumbersome calculations using this geometric idea. $\endgroup$ – t.b. Jul 10 '11 at 1:34
  • $\begingroup$ Thank you so much for pointing out this beautiful idea about the Circle of Apollonius. I didn't know about it. I am learning more about it now. $\endgroup$ – user786 Jul 10 '11 at 2:10
  • $\begingroup$ @Theo - Is it possible to simplify this answer? $\phi^{-1}(1/3) = 2$ so the condition $|f(2) - 1|/|f(2) + 1| = 1/3$ is saying $|\phi(f(\phi^{-1}(1/3)))| = 1/3$. So $|g(1/3)| = 1/3$ and you can skip the discussion about circles. Or have I made another mistake? $\endgroup$ – Sam Nead Jul 10 '11 at 19:11
  • $\begingroup$ @Sam: Sure, that works fine, but I wanted to explain what the condition means and how to see geometrically what is going on. See also the edit. $\endgroup$ – t.b. Jul 10 '11 at 19:15

EDIT - This answer is off-base - I misread upper half plane when right half plane was the hypothesis.


You should assume that $f$ is injective on the upper half plane. In that case $f$ will be a linear fractional transformation and you can make progress. If you don't assume injectivity then the function $f(z) = 1 + 2(z-1) - (z-1)^2$ is a counterexample.

Ooops! Even if you assume injectivity, consider the linear fractional transformation $$ f(z) = (z+2)/(-z+4) $$ having fixed points at $z = 1, 2$. This satisfies all hypotheses (injective, preserves the upper half plane, fixes $1$, and $|f(2)-1|/|f(2)+1| = 1/3$). However $f$ cannot be written in the form $(az+b)/(bz+a)$ because $f$ doesn't fix $z = -1$.

  • $\begingroup$ Ah this is true; $-1$ is also a fixed point. However $2$ need not be a fixed. Anyway if it were we would necessarily get the identity. $\endgroup$ – user786 Jul 9 '11 at 23:59
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    $\begingroup$ He needs the right half plane preserved not the upper half plane. $\endgroup$ – Zarrax Jul 10 '11 at 0:12
  • $\begingroup$ Here's my answer, although I feel that there is a logical mistake. We know that $1$ is a fixed point and also $-1$ has to be a fixed point if we want to find this Möbius transformation. So let $f(z)=\frac{az+b}{cz+d}$. Then the fixed point imply that we need to have $a+b-c-d=0$ and $a-b+c-d=0$ which implies that $a=d$ and $b=c$. So $f(z)=\frac{az+b}{bz+a}$ So there exists a $k$ such that $f(z)=\frac{z+k}{kz+1}$. With the other assumption on $f(2)$ we get $f(2)=\frac{3+e^{i\theta}}{3-e^{i\theta}}=\frac{2+k}{2k+1}$. We deduce that $k=\frac{1-e^{i\theta}}{1+e^{i\theta}}$. $\endgroup$ – user786 Jul 10 '11 at 0:41
  • $\begingroup$ So we end up with $f(z)=\frac{z+\frac{1-e^{i\theta}}{1+e^{i\theta}}}{\frac{1-e^{i\theta}}{1+e^{i\theta}}z+1}$ or after simplification, the desired form. The only problem I have is that I don't know that $-1$ is a fixed point. Can this be justified? I will do the exercise with Zarrax's hint now. $\endgroup$ – user786 Jul 10 '11 at 0:41
  • $\begingroup$ @user786 - Correct - you don't know -1 is fixed, and that seems to require Schwarz's lemma. The reason why I was talking about -1 being a fixed point was that I was producing a counterexample (where I get to assume all of the conclusions, as well as all of the hypotheses!). $\endgroup$ – Sam Nead Jul 10 '11 at 19:15

I got this for you to start with : if $|f(2)-1| = \frac 13 |f(2)+1|$, this means that $f(2)-1 = \frac{e^{i \theta}}{3} (f(2) + 1)$, which leads you to $$ f(2) = \frac{3+e^{i\theta}}{3-e^{i \theta}}. $$ I hope you can get something out of this.

  • $\begingroup$ Yes indeed, I wrote this at one point, I just didn't know what to do with it. $\endgroup$ – user786 Jul 9 '11 at 23:28
  • $\begingroup$ That would imply f(2) = 2 and the map would be the identity? Or f(2) = 1/2 and the map would look like... something else. I haven't done much complex analysis, I just pointed out something to try to help. The only thing I know about mobius transforms is that they are characterized by their values on three points and that they can be seen as stereographical projections. $\endgroup$ – Patrick Da Silva Jul 9 '11 at 23:38
  • $\begingroup$ @Patrick: A non-identity Moebius transformation can have at most 2 fixed points. It is the identity if it has more than 2 fixed points. $\endgroup$ – user786 Jul 9 '11 at 23:48
  • $\begingroup$ @Theo: What you say is true but we are not assuming that $f$ is a Möbius transformation; I only know that $f$ is holomorphic on the right half plane. I don't know its form. $\endgroup$ – user786 Jul 9 '11 at 23:51
  • $\begingroup$ In case you're interested, see my solution above. $\endgroup$ – t.b. Jul 10 '11 at 18:30

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