# Problem of the Week #50 - May 13th, 2013

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

-----

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 $f(z)\equiv z$.

-----

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

##### Well-known member
Staff member

Here's Ackbach's solution:
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.

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

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

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

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

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

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

That is,

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

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

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

Status
Not open for further replies.