MHB Can a Möbius transformation have more than two fixed points?

[Chris L T521]

Here's this week's problem.

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Problem: A fixed point of a function $f(z)$ is a point $z_0$ satisfying $f(z_0)=z_0$. Show that a Möbius transformation $f(z)$ can have at most two fixed points in the complex plane unless $f(z)\equiv z$.

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Recall: A Möbius transformation (also called a linear fractional transformation) is any function of the form $f(z)=\dfrac{az+b}{cz+d}$ with the restriction that $ad\neq bc$ (so that $f(z)$ is not a constant function).

 

[Chris L T521]

This week's problem was correctly answered by Ackbach and TheBigBadBen.

Here's Ackbach's solution:

Spoiler

It is known that Möbius transformations map circles to circles, and here we can think of straight lines as circles on the Riemann sphere. Also, recall that three points determine a circle. Therefore, if a Möbius transformation had three fixed points, it would have to have all points fixed, and would thus be the identity transformation. So, either a Möbius transformation has at most two fixed points, or it is the identity.

Here's TheBigBadBen's solution:

Spoiler

Suppose that $f$ is a Möbius transformation with three distinct fixed points.

We note that the Möbius transformation preserves cross-ratios. That is,

$$\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)} =\frac{(w_1-w_3)(w_2-w_4)}{(w_2-w_3)(w_1-w_4)}. $$

where $$w_k = f(z_k)$$ for $$k = 1,..,4$$.

Now, let $$z_1,z_2,z_3$$ be the fixed points under $f$. It follows that for an arbitrary $$z\in\mathbb{C}$$, we say that for $$w = f(z)$$ we have:

$$\frac{(z_1-z_3)(z_2-z)}{(z_2-z_3)(z_1-z)} =\frac{(z_1-z_3)(z_2-w)}{(z_2-z_3)(z_1-w)}. $$

That is,

$$\frac{(z_2-z)}{(z_1-z)} =\frac{(z_2-w)}{(z_1-w)}. $$

We note that $$g(z) = \frac{(z_2-z)}{(z_1-z)}$$ is injective for arbitrary values $$z_1≠z_2$$. It follows that $$w = z$$.

That is, if $f$ is a Möbius three distinct fixed points, then $f$ must be the constant function.