# Complex Analysis problem

1. Sep 6, 2014

### Jorriss

I came across an interesting problem that I have made no progress on.

Let f be an analytic function on the disc $D = \{z \in C ~|~ |z| < 1\}$ satisfying $f(0) = 1$. Is the following
statement true or false? If $f(a) = f^\prime(a)$ whenever $\frac{1+a}{a}$ and $\frac{1-a}{a}$ are prime numbers then $f(z) = e^z$ for all $z \in D$.

Obviously I know that $f(z) = e^z$ solves $f^\prime = f$, but I don't see how to use that here.

Last edited by a moderator: Sep 6, 2014
2. Sep 7, 2014

### mathwonk

i guess you want to show your function also satisfies f-f' = 0. since f is analytic, it suffices to show it satisfies this equation on an infinite set with a limit point in the open disc. do you know the principle of analytic continuation?

3. Sep 7, 2014

### WWGD

And notice the a here are all on the Real line.

4. Sep 7, 2014

### Infrared

Are you sure this is a serious problem? It looks to be equivalent to the twin prime conjecture.

Note that if $\frac{1+a}{a}=p$, then $a=\frac{1}{p-1}$, which is inside your domain for all odd primes $p$. $f$ will necessarily be the exponential iff there are infinitely many such values of a (so that they will have a limit point in the domain. We don't need to worry about the limit point being on the boundary $|z|=1$ since $a$ decreases as $p$ increases), but $\frac{1+a}{a}$ and $\frac{1-a}{a}$ are twin primes so your statement is equivalent to the twin prime conjecture.

5. Sep 8, 2014

### Jorriss

As it would turn out, no it wasn't. It was the first problem in a joke qualifying exam. The other problems made it far more obvious that it was not serious. Anyhow, thanks for your insight into the joke I suppose!

6. Sep 8, 2014

### WWGD

Kind of embarrassed at how I missed that. I guess I should read problems more carefully before trying to solve them.