Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f(z), g(z)$ are two entire functions, both have no zeros in the closed upper half plane. What does it mean/imply that $$\bigg| \lim_{y\rightarrow \infty}\frac{f(z)}{g(z)}\bigg|=c$$ ($z=x+iy$) i.e. after taking the limit inside the modulus the resulting function -depends on x- have modulus c. (In fact what I have is like: $|\lim_{y\rightarrow \infty} (\dots)|=|..ce^{ix}|=c$)

(I think it implies that $|f(z)|\leq c|g(z)|$ for all $z$ in the upper half plane, is that correct, and if so how to prove it!)

Also, what does it mean/imply that

$$\bigg|\lim_{y\rightarrow 0}\frac{f(z)}{g(z)}\bigg|=d$$

EDIT: $c$ and $d$ are non zero.

share|improve this question
Do you know what real limits mean? –  Phira Nov 7 '11 at 20:52
I think it cannot be true (in general) that $|f(z)|\leq c|g(z)|$ (unless $|f(z)|=c|g(z)|$). –  Valerio Capraro Nov 7 '11 at 20:52
@Lindsey: Did you mean to ask "are there any interesting non-trivial consequences of these limits using the fact $f$,$g$ are analytic" instead of merely "what do these limits mean?" –  Hurkyl Nov 7 '11 at 22:40
Yes, I hope there is some consequences fot this! –  Lindsey Mk Nov 8 '11 at 0:14

2 Answers 2

For your second question: If $f(z)/g(z) = h(z)$, then $h(z)$ is meromorphic in the complex plane and has no poles or zeros in the closed upper half plane. Now $\lim_{y \to 0} h(x+iy) = h(x)$, so you're saying $|h(x)| = d$ for $x \in \mathbb R$. By a version of the Schwarz Reflection Principle, $h(\overline{z}) = d^2/\overline{h(z)}$ for all $z$: in particular $f$ and $g$ have no zeros at all. That implies $f(z) = e^{F(z)}$ and $g(z) = e^{G(z)}$ where $F$ and $G$ are entire functions, where $\Re(F(x) - G(x)) = \ln(d)$ for $x \in \mathbb R$.

share|improve this answer
Sorry I just accidentally posted a comment that I was working on, and don't know how to delete it ... –  mathstribble Nov 8 '11 at 11:44
I don't think that it is true that $f$ and $g$ must have no zeros at all; they might have the same zeros as each other in the lower half plane. (Indeed, take any function $f$ that has no zeros in the upper half plane, but does have some in the lower half plane, and consider $g=d\cdot f$.) However, your argument does show that $f(z)=d\cdot g(z) \cdot e^{i\theta(z)}$, where $\theta$ is an entire function that is real on the real axis. –  mathstribble Nov 8 '11 at 11:48

I would take it to mean $\left|\lim_{y\to\infty}\frac{f(x+iy)}{g(x+iy)}\right|=c$, which -- at least a priori -- is something that depends on $x$, but not on $y$, because $y$ is bound by the limit operator. That is, it is just a claim about the magnitude of a real limit $\lim_{y\to\infty}(\cdots y\cdots)$, where $(\cdots y\cdots)$ happens to involve an $x$ and some complex functions.

It is not equivalent to $|f(z)|\le c|g(z)|$. As a simple counterexample, take $f(z)=0$, $g(z)=z$, which satisfies your condition, but not the original, since $\lim\frac{f(z)}{g(z)}=0$ for all $x$.

A counterexample for the other direction is $f(z)=c(1+e^{iz-1})$, $g(z)=1$, where $\lim_{\Im z\to\infty} \frac{f}{g}=c$ but $|f(z)|$ can vary both above and below $c|g(z)|$ over the upper half-plane.

share|improve this answer
Oh sorry I wrote the limit in a differen way, I edited it! –  Lindsey Mk Nov 7 '11 at 21:21
Concerning your example, thats right, but c and d are supposed to be nonzero. Does it make any difference in this case? –  Lindsey Mk Nov 7 '11 at 21:28
My example works fine with nonzero $c$. It satisfies $|f(z)|\le c|g(z)|$ everywhere (except at 0), but does not satisfy $|\lim f/g|=c$ for any nonzero $c$. –  Henning Makholm Nov 7 '11 at 21:30
That's right, again. My question was in one direction: finite limit $\rightarrow$ $|f|\leq c|g|$ –  Lindsey Mk Nov 7 '11 at 22:16
It doesn't work in that direction either. Answer edited with new counterexample. –  Henning Makholm Nov 7 '11 at 22:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.