# Thread: Nice Complex Analysis Problem

1. ## Nice Complex Analysis Problem

As I said I was taking an exam more than a month ago on Complex Analyis. One of the optional problems was (if I remember correcty):

"Find all complex rational functions which have modulos 1 on the unit circle"*.

Note: I cannot help you with this problem it uses a theorem which I never learned.

*)I was told that this problem was a favorite problem to give as an examination problem to graduate students in a certain event (which I forgot the name of).

2. It sounds as if they are asking you to rediscover the theory of Blaschke products.

3. Originally Posted by Opalg
It sounds as if they are asking you to rediscover the theory of Blaschke products.
This is exactly the term a professor told me it is. Can you post a proof?

And do you think this is a too hard of a question for an exam?

4. Here's a bare outline of how the proof goes. The set G of all complex rational functions which have modulus 1 on the unit circle is a group (under pointwise multiplication). For each complex number a with |a|<1, the function $\displaystyle f_a(z) = \frac{z-a}{1-\bar{a}z}$ is in G.

Now suppose that f is in G. If f has no zeros in the unit ball then f must be constant. (That is the tricky part of the argument. According to the Wikipedia page that I linked to in my previous comment above, "this fact is a consequence of the maximum principle for harmonic functions, applied to the harmonic function log(|f(z)|)".)

If f does have zeros on the unit ball, call them a_1, a_2, ..., a_n (counted according to their multiplicity). Then $\displaystyle f/(f_{a_1}f_{a_2}\cdots f_{a_n})$ is a function in G with no zeros in the unit ball and is therefore constant. Hence $\displaystyle f = \lambda f_{a_1}f_{a_2}\cdots f_{a_n}$ for some number $\displaystyle \lambda$ with $\displaystyle |\lambda|=1$.

That's the structure theorem for functions in G. Whether or not it's fair to ask it as an exam question depends on the students' background, I guess. I would say it's a pretty tough assignment for any student who hasn't previously seen the result.

5. The quick outline of the solution to this problem was presented to me using Schwartz-Reflection Principle.