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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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up vote 3 down vote accepted

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

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