# analytic, fixed point

• Feb 19th 2010, 02:33 PM
eskimo343
analytic, fixed point
Let $\displaystyle f(z)$ be an analytic function mapping the unit disc $\displaystyle \mathbb{D}$ into itself. Suppose that $\displaystyle f(z)$ has two different fixed points in the unit disc $\displaystyle \mathbb{D}$, show that $\displaystyle f$ must be the identity function $\displaystyle f(z)=z$ for all $\displaystyle z \in \mathbb{D}$.

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

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.
• Feb 19th 2010, 11:46 PM
chisigma
If $\displaystyle f(*)$ is analytic inside the unit disk, then it can be developped in Taylor expansion in [for example...] $\displaystyle z_{0}=0$...

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

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

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

... where $\displaystyle \gamma$ 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 $\displaystyle 2\pi i$. So...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 20th 2010, 02:49 AM
Opalg
Alternatively, let the fixed points be $\displaystyle z_1$ and $\displaystyle z_2$, and let g be a conformal map from the disk to itself that takes $\displaystyle z_1$ to the origin. Then $\displaystyle g f g^{-1}$ is an analytic map from the disk to itself that fixes the origin and also the point $\displaystyle g(z_2)$. You should now be able to use Schwarz's lemma to show that $\displaystyle g f g^{-1}$ is the identity and hence so is f.
• Feb 20th 2010, 02:57 AM
Laurent
Quote:

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 $\displaystyle 2\pi i$.

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?
• Feb 20th 2010, 06:07 AM
bkarpuz
Quote:

Originally Posted by Laurent
Quote:

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 $\displaystyle 2\pi i$.

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 $\displaystyle \chi\sigma$ proves that every analytic function in the unit disc is constant. Nice and simple. (Rofl)
• Feb 20th 2010, 09:12 AM
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 $\displaystyle f(z)= z^{2n+1}$ are analytic and have two fixed point in $\displaystyle z=1$ and $\displaystyle z=-1$, both on the unit disc... (Headbang) very sorry!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 20th 2010, 09:59 AM
Opalg
Quote:

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 $\displaystyle f(z)= z^{2n+1}$ are analytic and have two fixed point in $\displaystyle z=1$ and $\displaystyle z=-1$, both on the unit disc... (Headbang) very sorry!...

That is correct if you take $\displaystyle \mathbb{D}$ to be the closed unit disk. But I think that the original result is correct if $\displaystyle \mathbb{D}$ is the open disk.