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Complex Analysis

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    For each of the following functions f(z), find f'(z) and identify the maximal region for which f(z) is analytic.

    1. [itex]f(z)=1/(z^2+1)[/itex]
    2. [itex]f(z)=e^{-1/z}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    1. [itex]f'(z)=\frac{-2z}{(z^2+1)^2}[/itex] <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

    f(z) is analytic everywhere but + or - i because f'(z) is undefined there.

    Is that a true stament or is the correct statement ... f(z) is analytic everywhere but + or - i because f(z) is undefined there.

    2. [itex]f'(z)=\frac{e^{-1/z}}{z^2}[/itex] <--this part is easy. I'm having difficulty being certain of the maximum region for analyticity. Here is my attempt.

    f(z) is analytic everywhere but 0 because f'(z) is undefined there. However, f(z) is analytic at infinity.

    Is that a true stament or is the correct statement ... f(z) is analytic everywhere but 0 because f(z) is undefined there. However, f(z) is analytic at infinity.
     
  2. jcsd
  3. Sep 15, 2013 #2
    The short answer is that you are correct.

    In the future the simplest way to approach these problems is to remember the definition of analytic:

    Definition: A function ##f(z)## is analytic at a point ##z_{o}## if ##lim_{z \rightarrow z_{o}} \frac{f(z) - f(z_{o})}{z - z_{o}} = lim_{h \rightarrow 0} \frac{f(z_{o} + h) - f(z_{o})}{h}##.


    The maximal region for which ##f(z)## is analytic will be the entire complex plane with any singularities removed (read: with the places it is undefined removed.

    For example, for your second function we can write:

    The function ##f(z) = e^{\frac{-1}{z}}## is analytic on ℂ - {0}.
     
  4. Sep 15, 2013 #3

    vela

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    The first statement is the one you want. There's a theorem that says if a function is complex differentiable at a point, it's analytic at that point. Just because a function exists at a point doesn't mean it's analytic there.
     
  5. Sep 16, 2013 #4
    I don't think so huh vela? Doesn't it have to be complex-differentiable in some disc centered at the point in order for it to be analytic at that point?
     
  6. Sep 16, 2013 #5

    vela

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    Yeah, you're right.
     
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