How can it be shown that a Möbius transformation can have at most two fixed points unless it is $f(z) = z$?

Question

Obviously the identity fixes all points, but why do Möbius transformations fix at most two?

I know that Möbius transformations map circles and lines to circles and lines, but how does that imply that no more than two points can be fixed?

Answer 1

Let $z \to \frac{az+b}{cz+d}$ $(a,b,c,d \in \mathbb{C}$, and $ z \not = \frac{-d}{c}$) be the Mobius transformotation. A point $z_{0}$ if fixed if and only if $\frac{az_{0}+b}{cz_{0}+d}=z_{0}$,so $cz_{0}^{2}+(d-a)z_{0}-b=0$. This is an equation of degree at most $2$, hence has at most two solutions!

Answer 2

Hint

Mobius transform is a function of the form $f(z)=\frac{az+b}{cz+d}$. So for fixed points you need to solve $f(z)=z$. This results in (a possibly) second degree equation in $z$ and that will have (at most) two roots.

Answer 3

The case that ∞ is a fixed point is not a possibility, because as long as c is nonzero ∞ will be sent to a/c (thus not a fixed point).

However, if c is in fact zero, then ∞ IS a fixed point. Any other fixed point must then come from the equation:

$w=(a/d)z+(b/d)$

(the equation takes this form because c=0, and so a and d must be nonzero)

...and this equation has a single solution.

Answer 4

Consider a Mobius transformation $$Tz=\frac{az+b}{cz+d}$$ where $a, b, d\in\Bbb{C}$ and $ad-bc\neq 0$

Let $T$ is not the identity map.

Suppose $z\in\Bbb{C}$ is a fixed point of $T$ i.e $Tz=z$

$\frac{az+b}{cz+d}=z$

Implies $cz^2+(d-a) z-b=0\tag{1}$

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$\color{blue}{\text{Case}: 1(c\neq 0)}$

$(1) $ is quadratic polynomial in $\Bbb{C}$, hence can have at most two distinct complex roots.Implies $T$ can have at most $2$ complex fixed points.

And $c\neq 0$ implies $T(\infty) =\frac{a}{c}$, hence $\infty$ is not a fixed point.

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$\color{blue}{\text{Case}: 2(c= 0)}$

$(1) \implies (d-a) z+b=0$

1. If $a\neq d$ then $T$ has exactly one complex fixed point $z=\frac{b}{a-d}$

2. If $a= d$ and $b\neq 0$ then $T$ has no complex fixed point.

3. If $a=d$ and $b=0$ then $Tz=z$ for all $z\in\Bbb{C}$ fixes every complex numbers.

And $c=0$ implies $T(\infty) =\infty$

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Conclusion: A non identity Mobius map $T$ can have at most two distinct fixed points in $\Bbb{C}\cup\{\infty\}$