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: Question regarding holomorphic functions

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Given a complex valued function [tex]f(z) = 1/z^2+1[/tex] show the area for which its holomorphic?

    2. Relevant equations

    I know that if [tex]f:\Omega \rightarrow \mathbb{C}[/tex] and [tex]z_0 \in \Omega[/tex]

    then [tex]f'(z_0) = \lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}[/tex]

    if the limit exists then f is holomorphic at the point [tex]z_0[/tex]....

    3. The attempt at a solution

    To show the area for which f is holomorphic isn't this simply to check if the definition above can be applied to every [tex]z_0[/tex] of f?? Or am I missing something here?

    where the two possiblites for [tex]z_0 = \pm i[/tex] or is it simply that f is holomorphic on the area [tex]\Omega - \{\pm i\}[/tex] ??

    Best Regards
    Susanne
     
    Last edited: May 2, 2010
  2. jcsd
  3. May 2, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    First, is that [tex](1/z^2)+ 1= \frac{1}{z^2}+ 1[/tex] or [tex]1/(z^2+ 1)= \frac{1}{z^2+ 1}[/itex]?

    The first is not defined at z= 0 and the second is not defined at z= i or z= -i.
     
  4. May 2, 2010 #3
    HallsofIvy its suppose to to be

    [tex]f(z) = \frac{1}{z^2+1}[/tex]

    Isn't the point being here that the holomorphic definition can be applied to every point in [tex]\Omega[/tex] except [tex]\pm i[/tex] ??
     
  5. May 2, 2010 #4
    By that I mean that according to the definition of holomorph then a function can only be called holomorphic iff its complex differentiable in all points...

    But since [tex]f'(z_0 = \pm i)[/tex] doesn't exist then as I understand the definition of Holomorphic functions that f is holomorphic [tex]\forall z_0 \in \mathbb{C} \setminus \{\pm i\}[/tex]

    Because [tex](\frac{f}{g}^{\prime})(z_0) = 0[/tex] then [tex]z_0 = \pm i[/tex] and hence the definition of Holomorphic doesn't apply then [tex]z_0 = \pm i[/tex]. Then the area for which f is holomorphic is [tex]\forall z_0 \in \mathbb{C} \setminus \{\pm i\}[/tex].. As I see it...

    where [tex]\Omega(z_0,r)[/tex] where r>0 with the exception above is the area for which f is holomorphic...

    How is that Hallsoftivy??
     
    Last edited: May 2, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook