# Thread: another hard complex analysis question

1. ## another hard complex analysis question

Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

Edgar

Let f be an entire function.
(a) Prove that if e^f is bounded then f is constant.
(b) Prove that if Re f is bounded then f is constant.
Hint: Use suitable exponentials.

2. Originally Posted by edgar davids
Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

Edgar

Let f be an entire function.
(a) Prove that if e^f is bounded then f is constant.
(b) Prove that if Re f is bounded then f is constant.
Hint: Use suitable exponentials.
What you need is Picard's little theorem, find it here.

a) If $e^{f(z)}$ is bounded then so is $|f(x)|$,
let $K>0$ be a bound for $|f(z)|$ then both:

$2Ke^{i\theta}$ and $3Ke^{i\theta}$

are not values taken by $f(z)$, hence by Picard's theorem is a constant.

b) Same as above let $K>0$ be a bound, then both:

$2K+i$ and $3K+i$

are not values taken by $f(z)$, hence by Picard's theorem is a constant.

RonL

3. ## Liouvilles theorem

This results follow easily from Liouville's theorem. An entire bounded function is contstant

4. Originally Posted by edgar davids
Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

Edgar

Let f be an entire function.
(a) Prove that if e^f is bounded then f is constant.
(b) Prove that if Re f is bounded then f is constant.
Hint: Use suitable exponentials.
Originally Posted by petemarkou
This results follow easily from Liouville's theorem. An entire bounded function is contstant
Liouville's theorem gives $e^f$ is constant, and you need a little additional argument (since exp is not injective). For instance, $0=(e^f)'=f' e^f$ hence $f'=0$.

And for b), use $|e^{f}|=e^{\Re f}$ to reduce to the first question.

5. Originally Posted by CaptainBlack
What you need is Picard's little theorem, find it here.

a) If $e^{f(z)}$ is bounded then so is $|f(x)|$
There's probably no point in noting this (considering the delay), but anyway: a priori no, if $e^f$ if bounded, then $\Re f$ is bounded, but we don't know about $f$. If $f$ is entire, then a posteriori it is constant hence bounded.