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$f(z), g(z)$ 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.

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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
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2 Answers 2

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.

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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
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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 $x \in \mathbb R$.

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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
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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
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