Fixed points of $\phi_a(z)$ 
Prove that $\phi_a(z)=\frac{a-z}{1-\bar az}$ , $0<|a|<1$ has exactly two fixed points ; one inside the unit disc and the other outside the unit disc.

Putting $\phi_a(z)=z$ I find that there are exactly two fixed points which are $$\frac{1\pm \sqrt{1-|a|^2}}{\bar a}.$$
I am unable to find out the second part.
Let $z_1=\frac{1- \sqrt{1-|a|^2}}{\bar a}$. From $0<|a|<1$ , we get $z_1<\frac{1}{\bar a}=\frac{a}{|a|^2}$. Then , $|z_1|<\frac{1}{|a|}$. But how I show that $|z_1|<1$ so that I can say that $z_1$ lies inside the circle ?

You are solving the quadratic
$$\def\abar{\overline a}\abar z^2-2z+a=0\ .$$
By the usual product-of-roots formula we have
$$z_1z_2=a\,/\,\abar$$
and so
$$|z_1||z_2|=|a\,/\,\abar|=1\ .$$
Therefore one of $z_1$ and $z_2$ has modulus less than $1$, the other greater.
(It could be that both have modulus equal to $1$, but it is not very hard to see that this is impossible.)
