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This looks almost too easy where did I go wrong? Complex Analysis.

  1. Aug 26, 2013 #1
    1. The problem statement, all variables and given/known data

    http://www.math.northwestern.edu/graduate/prelims/AnalysisPrelim2010FallFinalVersion.pdf

    Problem 2 of Part III.

    2. Relevant equations

    Complex Analysis.

    3. The attempt at a solution

    So, I think my proof is wrong (since I never used the fact that it was [itex]f^2[/itex]) as opposed to [itex]f[/itex]. So, could you point out at where?

    Since [itex]f[/itex] is meromorphic, it can only have poles as discontinuities. We'll argue by means of contradiction. Let [itex]a[/itex] be a pole of [itex]f[/itex]. Since meromorphic functions have isolated zeroes, there exists a small disk with border [itex]\gamma[/itex] around [itex]a[/itex] such that [itex]f[/itex] has no zeroes, nor other singulaties. In particular, this implies [itex]g := \frac{1}{f^2}[/itex] is analytic on said disk. Applying Runge's theorem, we can obtain a sequence of polynomials [itex]g_n[/itex] uniformly converging to [itex]g[/itex]. So, [tex] \int_{\gamma} g_n f^2 dz \rightarrow \int_{\gamma} gf^2 dz = length(\gamma) [/tex] But by the given,

    [tex]\int_{\gamma} g_n f^2 dz = 0[/tex] for all [itex]g_n[/itex], which is a contradiction.

    Am I just invoking a theorem too powerful?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 26, 2013 #2

    mfb

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    An integral over a closed loop with a constant function is 0, not the length of the loop. This does not lead to a contradiction.
     
  4. Aug 27, 2013 #3
    Oh wow. Shameful. Shameful, shameful, shameful. Been doing too much real stuff lately haha. How about this?

    By the integral condition, it suffices to only consider [itex]f[/itex] having simple poles as discontinuities. Wlog, we assume it has a pole at [itex]z = 0 [/itex] and choose a contour around it which only has that as the only discontinuity in the region it defines. Thus, by the residue theorem and the integral condition, we have that [itex] \lim_{ z \rightarrow 0} z f(z) = 0 [/itex]. Thus, there exists some small c such that [itex] |z| \leq c \rightarrow |z f(z)| < 1 [/itex]. The function [itex] g(z) = z f(z) [/itex] has only 0 as a discontinuity in previously-mentioned region, and it's removable, so it's analytic on the domain defined by the contour. For [itex] |z| < c[/itex] we have by the maximum modulus principle that [itex]|g|[/itex] attains a max value at [itex]|z| = c [/itex], so in particular, [itex]|f(z)| < 1/c [/itex], for [itex]|z| < c [/itex], hence the discontinuity is removable, so not a pole.
     
  5. Aug 28, 2013 #4

    mfb

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    I don't understand that reasoning.

    I would simplify this to "no pole of any order could satisfy the lim condition, therefore f has not a pole". But you have to get the lim condition first.
     
  6. Aug 28, 2013 #5
    Alright. So, by the integral condition and the residue theorem (in that order), we get the equalities

    [tex]0 = \int zf^2 dz = Res(zf^2,0)[/tex]

    Since we are assuming it's a simple pole of [itex]f[/itex], it's an order 2 pole of [itex]f^2[/itex], hence simple pole of [itex]zf^2[/itex], so the residue boils down to the limit condition [itex] \lim_{z \rightarrow z} z(z f^2 (z) ) = 0 [/itex] which is equivalent to the stated limit condition before by taking a square root.

    We can assume it's a simple pole since if it was any higher, say [itex]m[/itex], we can simply replace [itex]f[/itex] with [itex]z^{m-1} f [/itex], which would then have a simple pole at [itex]z= 0[/itex] , then show it's not really a pole by the argument I produced earlier which means it couldn't have been a pole of said order for [itex]f[/itex].

    I tried to reduce to the simple pole case since for higher order, the residue involves taking derivatives, and that looked scary.
     
    Last edited: Aug 28, 2013
  7. Aug 29, 2013 #6

    mfb

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    Better start with modified polynomials directly, otherwise your assumption is not valid - your other cases do not look like the case you considered.
     
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