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Complex Analysis problem

  1. Sep 6, 2014 #1
    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. jcsd
  3. Sep 7, 2014 #2


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    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?
  4. Sep 7, 2014 #3


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    And notice the a here are all on the Real line.
  5. Sep 7, 2014 #4
    Are you sure this is a serious problem? It looks to be equivalent to the twin prime conjecture.

    Note that if [itex] \frac{1+a}{a}=p [/itex], then [itex] a=\frac{1}{p-1} [/itex], which is inside your domain for all odd primes [itex] p [/itex]. [itex] f [/itex] 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 [itex] |z|=1 [/itex] since [itex] a [/itex] decreases as [itex] p [/itex] increases), but [itex] \frac{1+a}{a} [/itex] and [itex] \frac{1-a}{a} [/itex] are twin primes so your statement is equivalent to the twin prime conjecture.
  6. Sep 8, 2014 #5
    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!
  7. Sep 8, 2014 #6


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