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The question asks:

Let $\varphi$ be a linear fractional transformation which maps the upper half plane $\{z : \operatorname{Im}(z) > 0\}$ onto itself. Prove that if there exist distinct $z_1$ and $z_2$ having positive imaginary parts with $\varphi(z_1) = z_1$ and $\varphi(z2) = z_2$ then $\varphi(z) = z$ for all $z$.

There is an apparently related question that asks: 2.) Let $\varphi$ be a linear fractional transformation which maps the unit disk $\{z : |z| < 1\}$ onto itself. Prove that if there exist distinct $z_1$ and $z_2$ in the disk with $\varphi(z_1)=z_1$ and $\varphi(z_2)=z_2$ then $\varphi(z)=z$ for all $z$.

I was wondering if there is some sort of easy approach to this type of question that I'm missing? I don't believe they were intended to be particularly difficult, but after a significant amount of time I've run out of good ideas.

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    Note the solutions to the quadratic equation with real coefficients are complex conjugates.2011-07-06

2 Answers 2

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Think symmetry. Since $\varphi$ takes the upper half-plane into itself, $\varphi(\bar z)=\overline{\varphi(z)}$. Since $z_1$ and $z_2$ are fixed,so are $\bar z_1$ and $\bar z_2$.

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Answer to your second question:

Every Möbius transformation that preserves the unit disk is of the form

$m(z) = e^{i\theta} \frac{a-z } {1 - \bar{a}z }$, where $a$ is a point in the unit disk.

Proof (à la Julián Aguirre) : Take $a$ to be point that gets mapped to the origin, so immediately one has that its inverse is mapped to infinity. Then you require that some point on the unit circle, say $1$ be mapped to another point on the unit circle (The boundary of the unit disk). Done.

Can you deduce from here that if such a Möbius Transformation fixes two points, it is the identity transformation?

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    In general, if $G$ is$a$group, $h$ is conjugate to $k$ if $k= ghg^{-1}$. Now let $h = \phi$ and let $g:\mathbb{D} \to \mathbb{H}$ be the Cayley transform. Then you get a Möbius transformation $k: \mathbb{H} \to \mathbb{H}$ having two fixed points in the upper half plane, and the problem is reduced to the one which is already settled.2011-07-07