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

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Chris L T521

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Jan 26, 2012
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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).

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by Ackbach and TheBigBadBen.

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.

Here's TheBigBadBen's solution:
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.
 
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