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I'm studying some complex analysis in preparation for a qualifier exam and I'm doing exercise $6.12$ from Robert Greene and Steven Krantz' book Function Theory of One Complex Variable.

I have $\Omega$ a simply connected domain in $\mathbb{C}$, with $P,Q \in \Omega$ two different points. $\phi_1 : \Omega \to \Omega$ and $\phi_2 : \Omega \to \Omega$ are conformal maps (that is, bijective and holomorphic) such that $\phi_1(P) = \phi_2(P)$ and $\phi_1(Q) = \phi_2(Q)$. Then the question is to prove that $\phi_1 \equiv \phi_2$.

I don't know why but I've been thinking about how to approach this without success for quite some time now. I've thought of maybe using the Riemann mapping theorem to say that there's a conformal map $\phi:\Omega \to \mathbb{D}$ to the open unit disk, and similar things but I'm not getting anywhere really.

Thus I would really appreciate some help with this problem, maybe some hints on how to proceed would be very appreciated. Thanks in advance for any help.

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I think you're on the right track. $\Omega$ is either all of $\mathbf{C}$ or it is conformal to the unit disk. You can write all of the automorphisms of the plane or disk down (Greene and Krantz do this, IIRC) and in either case there are two parameters; fixing two points nails these down. – Dylan Moreland Dec 5 '11 at 8:16
Oh I see, thank you very much @Dylan. – Adrián Barquero Dec 5 '11 at 8:23
up vote 8 down vote accepted

You are on the right track! You have to use the Riemann mapping theorem, and the fact that all conformal maps $D \to D$ are of the form $$z \mapsto e^{i\phi}\frac{z+b}{\overline{b}z+1},\: \: \phi \in \mathbf{R},\: \: |b|<1.$$ (This is essentially the group $\mathrm{PSL}_2(\mathbb{R})$ in disguise, since the unit disc and the upper half-plane are conformally equivalent and the automorphism group of the latter is $\mathrm{PSL}_2(\mathbb{R})$, which is perhaps easier to see.)

If $\sigma : \Omega \to D$ is a the conformal map, then $\sigma\phi_1\sigma^{-1}$ and $\sigma\phi_2\sigma^{-1}$ are maps $D \to D$ which coincide at two different points. Now you can use the characterization above to show that they must be equal.

Addendum: if $\Omega=\mathbf{C}$, then any holomorphic bijective map $\Omega \to \Omega$ is of the form $z \mapsto az+b$, and the statement is true also.

Addendum #2: Here's the complete solution as hinted to in the comments. By the above, we are left proving it for the conformal map $z \mapsto e^{i\phi}\frac{z+b}{\overline{b}z+1},\: \: \phi \in \mathbf{R},\: \: |b|<1$. Its fixed points are the roots of the quadratic polynomial $$z^2+z(1-e^{i\phi})/\overline{b}-e^{i\phi}\overline{b}/{\overline{b}}.$$

Since the product of the roots is the constant term $e^{i\phi}\overline{b}/{\overline{b}}$, which has modulus $1$, at most one of the roots can have modulus $<1$.

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Thank you very much Bruno. – Adrián Barquero Dec 5 '11 at 8:24
Don't forget the case where $\Omega$ is the whole complex plane. – Robert Israel Dec 5 '11 at 8:24
You are very welcome Adrián! – Bruno Joyal Dec 5 '11 at 8:24
@AdriánBarquero Show that a conformal map $D \to D$ as above has at most one fixed point inside $D$. Then if $\phi_1, \phi_2 : D \to D$ coincide at two points inside $D$, $\phi_1\phi_2^{-1}$ has two fixed points inside $D$. – Bruno Joyal Dec 5 '11 at 20:03
@AdriánBarquero: Extra hint: the fixed points are the roots of $z^2 + z(1-e^{i\phi})/\overline{b} - e^{i\phi}b/\overline{b}$. The product of the two fixed points is $- e^{i\phi}b/\overline{b}$, which has modulus $1$... – Bruno Joyal Dec 5 '11 at 20:06

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