R(z) be a rational function satisfies|R(z)|=1 when |z|=1. Prove that R(z) can be written as $\ R(z)=cz^m\prod_{n=1}^k{\frac {z-a_n}{1-\bar a_nz}}$ R(z) be a rational function satisfies|R(z)|=1 when |z|=1. Prove that R(z) can be written as $\ R(z)=cz^m\prod_{n=1}^k{\frac {z-a_n}{1-\bar a_nz}}$, c is a constant and $0\lt|a_n|\ne 1$.
Hint:  try to prove that if $z_0$ is a zero of R(z) then $\frac 1{\bar z_0}$ is a pole of R(z).
 A: Hints (beyond what you already have been given):
1) $R(z)$ has only finitely many zeroes and poles in $|z| < 1$ since $|R(z)| = 1$ when $|z| = 1$. 
2) Each ${\displaystyle {z - a_n \over 1 - \bar{a_n} z}}$ has a zero at $z = a_n$ if $|a_n| < 1$ and a pole at ${1 \over \bar{a_n}}$ if $|a_n| > 1$. It also has magnitude $1$ when $|z| = 1$.
A: This follows from a categorization theorem of Blaschke. Essentially, every analytic function that maps the unit disc onto itself, must be a Blaschke product. 
Preliminary exercises to this include the following. 
1) Let $f:\bar{D}\to \bar{D}$ is analytic where $|f(z)|=1$ for $|z|=1$. Show that if $f$ has no zeros in $D$ then $f$ must be a constant. (Use maximum modulus principle).
2) Let $f:\bar{D}\to\bar{D}$ be an analytic function where $f(0)=0$ and $|f(z)|=1$ for $|z|=1$. Then $f(z)=\alpha z$ for some $\alpha\in\mathbb{C}$. (Schwarz Lemma is a special case of Blaschke's result) 
3) For some $|w|<1$, define a Blaschke factor, $$f(z)=\frac{z-w}{1-\bar{w}z}.$$ Show that $f(z)$ maps $\bar{D}\to\bar{D}$ and that $|f(z)|=1$ for every $|z|=1$. 
4) Suppose $f:\bar{D}\to\bar{D}$ is analytic with $|f(z)|=1$ and $|z|=1$. By the uniqueness theorem $f$ has finitely many zeros in $D$, $\{a_1,\ldots,a_n\}$. Show that $f$ is a Blaschke product of it's zeros, $$f(z)=c\prod_{k=1}^n \frac{z-a_k}{1-\bar{a_k}z}.$$
To prove (4), we do the following, define, $$h(z)=f(z)\prod_{k=1}^n \frac{1-\bar{a_k}z}{z-a_k}.$$ Certainly, $h(z)$ is analytic on $D$ and has no zeros (since we canceled them). By (3), $h:\bar{D}\to\bar{D}$ and $|h(z)|=1$ when $|z|=1$. Thus, we conclude that $h(z)$ must be a constant by (1). Hence, $f(z)$ is indeed a Blaschke product. 
I am sure a slight modification of this trick will solve your problem. 
