Prove f has at least one fixed point on the boundary Let $$f(z) = \frac{a-z^2}{1- \bar a \cdot z^2}$$ where $a \in D=\{|z| <1\} $.
Denote the boundary as $S =\{|z| =1\} $. Show that $f$ has at least one fixed point $w \in S$.
Obviously, $f(S) \subseteq S$, so I can't compare $|f(z)|$ with $|z|$ in $S$. And if I simplify $f(z)-z$, I get $\bar a z^3 - z^2 -z +a$. I don't know how to use Rouche's theorem then. 
Also, Rouche's theorem only tells the number of roots inside the circle or annulus. I don't know how to find the root on the boundary. Maybe replace $z$ by $\frac{1}{z}$?
 A: You are trying to solve $f(z)=z$ and see a root on the unit circle.  This is just
$\frac{a-z^2}{1-\overline{a}z^2}=z$ or $$\overline{a}z^3-z^2-z+a=0.$$
You want to see that it has a root on the unit circle.  
Multiplying the displayed equation through by $z^{-3}$ and taking the complex conjugate, we see that if $z$ solves the equation then so does $\frac{1}{\overline{z}}$. 
The map $T(z)=\frac{1}{\overline{z}}$ is an involution of the Riemann sphere that exchanges the unit disk minus $0$ with the complex numbers of norm greater than $1$. Notice $0$ is not a root.
If there are no roots on the unit circle, then there must be an even number of roots, as the involution above exchanges each root outside the unit circle with a root inside the unit circle. This contradicts the fact that a polynomial of degree three has three roots counting multiplicities.
A: Sorry I do not know much about complex number, but here is one using geometry.
Assume no fixed point.
Hence $\frac{f(z)}{z}$ is a continuous function that does not pass through $1$. Obviously, since $f(z)$ map $S$ to subset of $S$ we have $\frac{f(z)}{z}$ also map $S$ to subset of $S$.
We have $f(-z)=f(z)$ (since $z$ always get squared in the formula for $f(z)$). Now we have $\frac{f(-z)}{-z}=\frac{f(z)}{-z}=-\frac{f(z)}{z}$. In other word, if $w$ is in the range of $\frac{f(z)}{z}$, so is $-w$. Which means that: (a) $\frac{f(z)}{z}$, in addition to $1$, must not pass through $-1$ either; and (b) it also have to pass through 2 opposite points.
Now we have a circle. $\frac{f(z)}{z}$ cannot pass through neither $1$ or $-1$ but must be continuous and somehow contain in its range 2 opposite point. Contradiction.
Side remark: looks like the requirement that $|a|<1$ is nearly completely superfluous. The argument still work even if $|a|>1$, and if $|a|=1$ and $a\not=1$ then there are 2 singularity...except that they're completely removable, because both zeros of numerator and denominator are order 1, and the fixed point does not happen at that removable singularity. As for $a=1$, then $f(z)$ do not have a fixed point indeed. So I guess the condition that $|a|<1$ is only needed to rule out $a=1$.
A: Look at the function $g:\mathbb R \to \mathbb R$ such that $f(e^{ix}) = e^{ig(x)}$.  Clearly the function is only defined up to a multiple of $2\pi$, but you will notice that $g(2\pi) = g(0) + 4\pi$ if $g$ is continuous.  (The winding number of the curve $x \mapsto f(e^{ix})$ is 2.)  So by messing around with the intermediate value theorem, you should be able to deduce the existence of a point $x_0$ such that $x_0 = g(x_0) \pmod {2\pi}$.
A: write
$f(z) = \frac{a-z^2}{1- \bar a \cdot z^2}=\frac{p(z)}{q(z)}$
$\gamma(t):=\exp\big(2\pi i \cdot t\big)$ for $t\in[0,1]$
e.g. Rouche's Theorem tells you that $n\big(p\circ\gamma,0\big)=2$ and $n\big(q\circ\gamma,0\big)=0$
$\implies n\big(f\circ\gamma,0\big)=2-0=2$
with the inversion map $h:z\mapsto z^{-1}$, consider the curve $\sigma$ given by
$\sigma(t) := \big(h\circ \gamma(t)\big) \cdot \big(f\circ\gamma(t)\big)$
$\implies n\big(\sigma,0\big)=n\big(h\circ\gamma, 0\big)+n\big(f\circ\gamma, 0\big)=-1+2=1\neq 0$
the non-zero winding number tells us $\frac{f}{z}:S^1\longrightarrow S^1$ is surjective
(if not surjective, we may rotate and stereographically project into $\mathbb R$ which is convex hence null homotopic)
In particular $z=1$ is in its image thus $\frac{f\big(\gamma(t')\big)}{\gamma(t')}=1$ for some $t'\in[0,1)$ i.e. $f\big(\gamma(t')\big) =\gamma(t')$ so $f$ has a fixed point.
remark:
if $\vert a\vert \gt 1$ then we'd have
$n\big(f\circ\gamma,0\big)=0-2=-2$ and $n\big(\sigma,0\big)=-1-2=-3$
which actually implies at least 3 distinct fixed points on the unit circle
