Rational functions with absolute value $1$ on $\mathbb{S}^1$ This is a question in Complex Analysis. It should not be too difficult but I am missing the trick.
The question asks us to find general form of rational functions $R:\mathbb{C}\to\mathbb{C}$ such that $|R(z)|=1$ for all $|z|=1$. It then asks us to find a relation between poles and zeros of such functions.
I have not made much progress. But I know all $z^{k} (k\in\mathbb{Z})$ satisfies the condition. Their zeros are $0$ (or $\infty$)  inverse to their poles $\infty$ (or $0$). Another direction I thought about was that if $R(z)$ is such a function then so is $R(1/z)$. 
Thanks very much! 
 A: Consider $ \frac{1}{\overline{R(\overline{1/z})}}$. This agrees with $R$ on the unit circle, so it equals $R$, because rational functions agreeing at infinitely many points are equal. What does this allow you to conclude? (Suppose $z_0$ is a zero of $R(z)$. Then something related is a pole for the expression above. But they're equal...)
A: This answer uses methods not yet developed when Ahlfors assigns the exercise in question. But I think it's conceptually clearer. (It's the method Jim Belk was alluding to above.)
Let $R(z)$ be as in the question. If $f(z)$ is an entire conformal map sending the unit circle to the real axis, we can form a function $Q(z)=f(R(f^{-1}(z)))$. Then $Q(z)$ takes only real values on the real line. Then $Q(z)=\overline{Q(\overline{z})}$ on the real line. Because they agree on a dense set, they must agree everywhere. You can then run some the same argument as in my other answer. (If $z_0$ is a zero of $Q(z)$, then $\bar z_0$ must also be, etc.) Finally, use your knowledge of conformal maps to pull what you discovered back to the unit disk. 
