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

[l'Hôpital]

Homework Statement

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

Problem 2 of Part III.

Homework Equations

Complex Analysis.

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?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[mfb]

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.

 

[l'Hôpital]

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.

 

[mfb]

Thus, by the residue theorem and the integral condition, we have that [itex] \lim_{ z \rightarrow 0} z f(z) = 0 [/itex].

I don't understand that reasoning.

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.

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.

 

[l'Hôpital]

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.

 

[mfb]

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].

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