# Show that if $z\in\mathbb{E}$ is a fixed point of $\phi\in \operatorname{Aut}(\mathbb{E})$, then so is $\frac{1}{\overline{z}}$

Show that if $z\in\mathbb{E}$ is a fixed point of $\phi\in \operatorname{Aut}(\mathbb{E})$, then so is $\frac{1}{\overline{z}}$, where $\mathbb{E}$ is the open unit disk.

According to my book all automorphisms from the open unit disk to itself are given by $\phi(z) = e^{i\theta}\frac{z-a}{\overline{a}z - 1}$ for $a\in\mathbb{E}$ and $0\leq\theta<2\pi$. First I tried simply setting that equal to z and then multiplying both sides by $\frac{1}{z\overline{z}}$ to obtain: $$e^{i\theta}\frac{1}{z\overline{z}}\frac{z-a}{\overline{a}z - 1} = \frac{1}{\overline{z}}$$

and then attempting to simplify the left side so that $\frac{1}{\overline{z}}$ takes the place of where z used to be, but I can't for the life of me seem to massage it algebraically into that form.

Next I turned it into a quadratic equation equal to zero and found its roots, but these roots depend on both $\theta$ and $a$ in a pretty complicated way and I'm unable to separate the roots out into their real and imaginary parts so that I can conjugate and invert and check for equality.

Now I'm thinking maybe there's some algebraic theorem that says under certain conditions that a quadratic equation over the complex numbers must have roots which are either inverted conjugates of each other or themselves, or something to that effect, but I don't know it..

Some help would be much appreciated, thanks.

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Please (i) Don't start a post mid-thought. The title is not part of the post. (ii) Don't use only the [homework] tag; please include a tag that indicates the subject matter as well. –  Arturo Magidin Apr 1 '12 at 20:54
Instead of writing x\;\epsilon\;y write x\in y, which gets the correct spacing. In actual LaTeX code, if you prefer the look of \epsilon to the default \in, you can change it. Here consistency among all posts is much more important than one's personal tastes about the look of specific characters! –  Mariano Suárez-Alvarez Apr 1 '12 at 20:58
Sorry about that, I figured just putting the premise as the title would be fine, I just wasn't sure all of it would fit. Ok cool I didn't even know about the \in command, thanks. –  Tyroshi pleasure barge Apr 1 '12 at 20:58
0% accept rank? Do you know about accepting answers to your questions, and about the importance of doing so? –  Gerry Myerson Apr 2 '12 at 0:39
No I have no idea what that is about, could you explain? –  Tyroshi pleasure barge Apr 2 '12 at 0:56

$\,z\in\Bbb E\,$ is a fixed point of $\,\phi(z)\,$ means $$e^{i\theta}\frac{z-a}{\overline az-1}=z\stackrel{\text{conjugate}}\Longrightarrow e^{-i\theta}\frac{\overline z-\overline a}{a\overline z-1}=\overline z\stackrel{\text{take inverses}}\Longrightarrow e^{i\theta}\frac{1-a\overline z}{\overline a-\overline z}=\frac{1}{\overline z}\stackrel{\text{in LHS factor out}\,\overline z}\Longrightarrow$$ $$\Longrightarrow \phi\left(\frac{1}{\overline z}\right):=e^{i\theta}\frac{\frac{1}{\overline z}-a}{\frac{\overline a}{\overline z}-1}=\frac{1}{\overline z}\Longrightarrow \,\,\frac{1}{\overline z}\,\,\text{is a fixed point of}\,\,\phi(z)$$