Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex Analysis-Analytic Functions

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove or find a proper counterexample:
    There exists an analytic function f(z) in a pierced neighborhood of z=0 (i.e a neighborhood of z=0 , which doesn't contain z=0) which satisfies: [tex] f^3 (z) = z^2 [/tex].

    2. Relevant equations
    3. The attempt at a solution
    Got no clue...

    Hope you'll be able to help me


    Thanks in advance
     
  2. jcsd
  3. Jun 14, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    If, by [itex]f^3(z)[/itex], you mean [itex](f(x))^3[/itex] and not [itex]f(f(f(x)))[/itex], then all you need to do is show that [itex]f(x)= z^{2/3}[/itex] is analytic everywhere except at z= 0.
     
  4. Jun 14, 2010 #3
    Actually, it's excatly what I mean, but why the function you gave isn't analytic at 0?
    It's actually an entire function...isn't it?

    Thanks
     
  5. Jun 14, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    I didn't say it wasn't- but your problem was to determine if "There exists an analytic function f(z) in a pierced neighborhood of z=0 (i.e a neighborhood of z=0 , which doesn't contain z=0)".

    If such a function is analytic everywhere it is certainly analytic in that neighborhood!

    But what make you so certain [itex]z^{2/3}[/itex] is analytic at z= 0? It's derivative does not exist there!
     
  6. Jun 14, 2010 #5
    I think the problem here is to show that you can't have such a function, the best you can do is cut away a branch cut line. Just omitting the point z = 0 is not enough.

    I think the proof has to use the integral of the derivative of the function and then show that this integral around the point z = 0 is not zero.
     
  7. Jun 15, 2010 #6
    Thanks ... I'll try this...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook