- Thread starter
- Moderator
- #1
- Jan 26, 2012
- 995
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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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!