# Math Help - analytic, fixed point

1. ## analytic, fixed point

Let $f(z)$ be an analytic function mapping the unit disc $\mathbb{D}$ into itself. Suppose that $f(z)$ has two different fixed points in the unit disc $\mathbb{D}$, show that $f$ must be the identity function $f(z)=z$ for all $z \in \mathbb{D}$.

$z_0$ is called a fixed point for a function $f$, if $f(z_0)=z_0$.

For this one, I was thinking to use Rouché's Theorem or Schwarz lemma. However, we are dealing with fixed points here. So, I don't see what to do right now. I need a few hints on this one. Thanks.

2. If $f(*)$ is analytic inside the unit disk, then it can be developped in Taylor expansion in [for example...] $z_{0}=0$...

$f(z)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\cdot z^{n}$ (1)

... where the $f^{n} (0)$ are given by the Cauchy integrals...

$f^{n} (0) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{z^{n+1}}\cdot dz$ (1)

... where $\gamma$ is the unit circle. You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to $2\pi i$. So...

Kind regards

$\chi$ $\sigma$

3. Alternatively, let the fixed points be $z_1$ and $z_2$, and let g be a conformal map from the disk to itself that takes $z_1$ to the origin. Then $g f g^{-1}$ is an analytic map from the disk to itself that fixes the origin and also the point $g(z_2)$. You should now be able to use Schwarz's lemma to show that $g f g^{-1}$ is the identity and hence so is f.

4. Originally Posted by chisigma
You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to $2\pi i$.
I may be missing something obvious, but I can't see how to prove that. How do you use the two fixed point to compute the integral?

5. Originally Posted by Laurent
Originally Posted by chisigma
You can verify that the line integral in (2) vanishes forall n with the only exception of n=1 where is is equal to $2\pi i$.
I may be missing something obvious, but I can't see how to prove that. How do you use the two fixed point to compute the integral?
Actually $\chi\sigma$ proves that every analytic function in the unit disc is constant. Nice and simple.

6. It is evident that I read hurriedly the post of eskimo 343... with a little more carefull reading it is not difficult to 'discover' that there is an evident counterexample: all the complex functions of the form $f(z)= z^{2n+1}$ are analytic and have two fixed point in $z=1$ and $z=-1$, both on the unit disc... very sorry!...

Kind regards

$\chi$ $\sigma$

7. Originally Posted by chisigma
It is evident that I read hurriedly the post of eskimo 343... with a little more carefull reading it is not difficult to 'discover' that there is an evident counterexample: all the complex functions of the form $f(z)= z^{2n+1}$ are analytic and have two fixed point in $z=1$ and $z=-1$, both on the unit disc... very sorry!...
That is correct if you take $\mathbb{D}$ to be the closed unit disk. But I think that the original result is correct if $\mathbb{D}$ is the open disk.