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 f^2) as opposed to f. So, could you point out at where?

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

\int_{\gamma} g_n f^2 dz = 0 for all g_n, 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 f having simple poles as discontinuities. Wlog, we assume it has a pole at z = 0 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 \lim_{ z \rightarrow 0} z f(z) = 0. Thus, there exists some small c such that |z| \leq c \rightarrow |z f(z)| < 1. The function g(z) = z f(z) 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 |z| < c we have by the maximum modulus principle that |g| attains a max value at |z| = c, so in particular, |f(z)| < 1/c, for |z| < c, hence the discontinuity is removable, so not a pole.

 

[mfb]

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

I don't understand that reasoning.

Thus, there exists some small c such that |z| \leq c \rightarrow |z f(z)| < 1. The function g(z) = z f(z) 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 |z| < c we have by the maximum modulus principle that |g| attains a max value at |z| = c, so in particular, |f(z)| < 1/c, for |z| < c, 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

0 = \int zf^2 dz = Res(zf^2,0)

Since we are assuming it's a simple pole of f, it's an order 2 pole of f^2, hence simple pole of zf^2, so the residue boils down to the limit condition \lim_{z \rightarrow z} z(z f^2 (z) ) = 0 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 m, we can simply replace f with z^{m-1} f, which would then have a simple pole at z= 0 , 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 f.

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 m, we can simply replace f with z^{m-1} f, which would then have a simple pole at z= 0 , 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 f.

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