Complex analysis - Louvilles Theorem (I think)

[jrp131191]

Hi I am doing a past exam for my complex analysis course and I should just mention right now that while it's a mix of pure/applied math I have never done a pure math unit before and i really really really suck at doing proofs and such..

Given c>0 and f(z) is entire such that |f(z)| ≤ c|z| show that :

f(z)=wz for some complex constant w.

I just have no idea how to tackle problems like this whatsoever and have to turn to google and notes which I obviously won't be able to do in the exam. Also I have trouble remembering all these theorems, corollaries, propositions..

My attempt at a solution was to state Louvilles theorem which is that if:

|f(z)|≤M and f(z) is entire then f(z)=w.. I don't really know where to go from here..

Any tips for tackling problems like this would be really appreciated!

 

[alberto7]

I think you are in the good path. Just apply Liouville's theorem to the auxiliary function $$g(z)=\frac{f(z)}{z}\text,$$ which is entire because $f(z)=0$ and we can continue $g$ to the origin as $$g(0)=\lim_{z\to 0}\frac{f(z)}{z}\text.$$

 

[Bacle2]

I think you are in the good path. Just apply Liouville's theorem to the auxiliary function $$g(z)=\frac{f(z)}{z}\text,$$ which is entire because $f(z)=0$ and we can continue $g$ to the origin as $$g(0)=\lim_{z\to 0}\frac{f(z)}{z}\text.$$

Alberto7: what do you mean f(z)=0? Maybe you mean to say that, _if_ f(z)=0 , then

the discontinuity at z=0 can be removed (e.g., by the Riemann removable singularity

theorem, since |f(z)/z|<c <oo)?

Then, following (what I think was) Alberto7's idea ,

|f(z)/z| itself is entire and bounded, so...

 

[alberto7]

Sorry, I meant $f(0)=0$.

 

[Erland]

jrp: you should have some theorem in your textbook saying that if f is analytic (holomorphic) and bounded in some punctured neigborhood of a point a (or something simililar), then f can be extended to be analytic at a too. Apply this to f(z)/z and proceed as Alberto7 and Bacle2 suggested.