Recent content by brian87

Well, it seems that I got lost somewhere from post 4 onwards. (something about branch cuts) Well, Dick, what's the part that does not make sense with the integral method? [I note that although I know the series representation of f, this question was in a chapter before series were introduced...

I've looked through the proof of Liouville's theorem, it's just that I can't get that implication that such a M exists that |f(z)| < M for all z. (the earlier method i tried above did not use anything from the proof of liouville's theorem)

Ah, would this then work? By Cauchy's Integral Theorem, since f is entire, we have: f(z0) = 1/(2*pi*i) \oint\frac{f(z)}{z-z0}dz But since we have \lim_{z \to \infty}\frac{f(z)}{z} = 0, pick any circle with R -> infinity, thus we have f(z0) to be 0 everywhere which yields f(z) = constant

Whoops, my mistake then But really, can someone explain to me how does the condition that f is entire and lim z -> infinity f(z)/z = 0 imply that f is bounded? Thanks :) [edit: It might be helpful to note that in the course I'm taking, we're doing really basic complex analysis, nothing on the...

Would this solution be sufficient? f is entire, then it has a Taylor representation about 0: f(z) = a0 + a1*z + a2*z^2 + ... Also, we have lim z-> infinity f(z)/z = 0, i.e. we must have a1, a2, ... = 0 thus f(z) = a0, a constant. I used some reasoning from real analysis, can this be...

So genneth, is my previous statement true? Thanks

Hi, does this mean that the lack of singularities for finite z and it being finite at infinity imply that it's bounded on the complex plane? Thanks (Just looking for a clarification here)

Homework Statement Suppose that f is entire and \lim_{z \to \infty}\frac{f(z)}{z} = 0. Prove that f is constant. z, and f(z) are in the complex plane The attempt at a solution I've tried to find out how the condition \lim_{z \to \infty}\frac{f(z)}{z} = 0 implies the function itself is...