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.