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Complex Analysis Proof of Constant Function

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that if a function f:c->c is analytic and lim as z to infinity of f(z)/z = 0 that f is constant.

    2. Relevant equations
    Cauchy Integral Formula for the first derivative (want to show this is 0 ie: constant)

    f prime (z) = 1/(2ipi)*Integral over alpha (circle radius r) f(x)/(x-z)^2 dx.

    3. The attempt at a solution

    So, I think I'm right in incorporating the cauchy integral formula for the first derivative into my proof. I hope that I can show that the integral over the closed curve is 0 for all z, meaning that 0 is the derivative of f and therefore f is constant.

    Problems: Not sure how to show this integral is 0, since using this equation the new integral has a critical point over the point z. I'm assuming that the introduction of the assumption that the limit of f(z)/z = 0 needs to be employed to take care of this.

    In other words, I believe I have equiped myself with the right tools, though I'm not sure how to start implamenting them, I'll update the post as I continue working on this problem though any help would be appreciated.
  2. jcsd
  3. Feb 20, 2007 #2
    Hey, I think I may have gotten another method of solving the problem. Though when I took the limit I was a little unsure if I was allowed to say its 0.


    Since f is analytic its derivative is analytic.

    f prime (0) = Lim z to 0 (f(z) - f(0))/(z-0) = Lim z to 0 (f(z)-f(0))/z

    Taking the limit on both sides we get: (Lim assumes z to infinity, Lim' is z to 0)

    Lim f'(0) = Lim' Lim f(z)/z - f(0)/z = Lim' Lim -f(0)/z
    Lim f'(0) = Lim' Lim -f(0)/z

    -Lim f(0)/z = 0 since f(0) is a fixed number.

    Thus Lim f'(0) = Lim' 0 = 0

    Since f'(z) is 0 at the origin and its domain (C) has an accumilation point (anything on the plane) the entire connected domain must also be 0. Because the entire domain is connected we can state that f'(z) is 0 everywhere and therefore the function f(z) is constant.

    Edit: Nvm, I can see that I cant commute the limits like I did. :*(

    Though, I may be able to use the same logic using the cauchy integral formula and show that the limit of the integral is 0 as x goes to infinity.

    Question: Integrals and limits commute right?
    Last edited: Feb 20, 2007
  4. Feb 21, 2007 #3

    Gib Z

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    Homework Helper

    Question: Integrals and limits commute right?

    Answer: That doesn't make sense.
  5. Feb 21, 2007 #4


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    Staff Emeritus
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    It does make sense but is not necessarily true! The whole question of when limits (and the integral itself involves a limit) commute is a major part of analysis.
  6. Feb 21, 2007 #5
    Our class hasnt covered (at all) cases in which limits and integrals commute (obviously by my question that is evident). Is the way I'm using *taking the limit of an integral* seem like the correct route to take or should I try finding another method to prove the claim. I'm a little lost on how to incorporate the limit of f(x)/x as x goes to infinity without taking the limit explicitley.

    The only idea I have right now is to show that f(x)/(x-z)^2 is bounded for all x. Though I have no idea how I can do this without knowing f(x). I can try using the limit to prove that it must be bounded but again I'm unsure how to extrapolate this information.

    Lim x to infinity of f(x)/x implies that for any epsilon > 0 I can find a S > 0 such that for all ABS(X) > S (Numbers outside the disk of radius S) it implies that ABS(f(x)/x) < Epsilon.

    So... it doesnt really tell me anything is bounded, only that the complex numbers outside of a growing disk grow arbitrarly close to 0. It doesnt say that anything inside the disk namely f(0) is bounded by anything, nor does it state these facts about the actual function in question f(x)/(x-z)^2.
    Last edited: Feb 21, 2007
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