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.
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.
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.