# Math Help - Entire Functions

1. ## Entire Functions

let f be an entire function..

1) prove if e^f is bounded then f is constant
2) prove that if Re f is bounded then f is constant

i'm guessing you would have to use suitable exponentials but i don't have a good enough idea of what to do here. any help would be greatly appreciated xx

2. Originally Posted by joanne_q
let f be an entire function..

1) prove if e^f is bounded then f is constant
If f is entire then so is e^f, so the result follows from Liouville's theorem.
Originally Posted by joanne_q
2) prove that if Re f is bounded then f is constant
$e^f = e^{\text{re}\,f}e^{i\,\text{im}\,f}$, so $|e^f| = e^{\text{re}\,f}$. If re f is bounded then so is e^f, and the result follows from 1).

3. Originally Posted by Opalg
If f is entire then so is e^f, so the result follows from Liouville's theorem.
how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???

4. Originally Posted by smoothman
how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
exactly what was on my mind too. any suggestions?

5. Originally Posted by smoothman
how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...
I suppose this just amounts to the chain rule from calculus. If f is differentiable then so is e^f, with $\frac d{dz}(e^{f(z)}) = f'(z)\,e^{f(z)}$. The definition of an entire function is that it is a function that is everywhere differentiable, and it follows that if f is entire then so is e^f.

Originally Posted by smoothman
also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
Liouville's theorem says that if an entire function is bounded then it is constant. If e^f is constant then so is f. You don't need to bring the Taylor series into it.

6. Originally Posted by joanne_q
exactly what was on my mind too. any suggestions?
Originally Posted by smoothman
how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
Are you in the same class?