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The following is exercise 10.20 in Rudin's R&CA.

Suppose $f \in H(U)$, $g \in H(U)$, and neither $f$ nor $g$ has a zero in $U$. If $$ \frac{f'}{f}\left(\frac{1}{n}\right) = \frac{g'}{g}\left(\frac{1}{n}\right) \quad (n = 1, 2, 3, \ldots) $$

Find another simple relation between $f$ and $g$.

EDIT: I didn't know $U$ is the unit disc when I posted this. I thought it could be any open set.

$0$ is a limit point of the zeros of the function $\varphi = f'/f - g'/g$. However, $0$ isn't necessarily in $U$. I'm not sure how to deal with it if $0$ is a pole or essential singularity of $\varphi$, and I couldn't find much info about limit points of zeros that are outside the domain of a holomorphic function. Ultimately, I suspect $\varphi = 0$ everywhere and thus $g = \lambda f$ for some constant $\lambda$.

Am I on the right track? Is there a better approach here? Thanks.

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What is $U$? Any open set? – mrf Jan 23 '13 at 23:28
@mrf This is all what the exercise says. It doesn't specify what $U$ is, so assume it can be any open set that contains $1, 1/2, \ldots$. – PeterM Jan 23 '13 at 23:30
I just browsed through my copy. It seems as $U$ in fact is the unit disc. Check p110 (at least in my edition, look for "unit disc" in the index). – mrf Jan 23 '13 at 23:42
@mrf Yes, Davide Giraudo just pointed this out. The exercise becomes very easy once you know this. I spent so much time on nothing. :( – PeterM Jan 23 '13 at 23:43
up vote 1 down vote accepted

Let $h:=f/g$ (well defined). Its derivative is $0$ and if $U$ is the open unit disk, we get $h'\equiv 0$ hence $f$ and $g$ are proportional.

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How do you know the derivative is $0$? We don't know if $U$ is the open disc. $f$ and $g$ may not be defined at $0$. – PeterM Jan 23 '13 at 23:17
In the previous exercises, $U$ is assumed to be the open unit disk. I guessed (but I may be wrong) that it's still the case in this exercise. – Davide Giraudo Jan 23 '13 at 23:35
Gosh you're right. And I've been researching this for 2 hours to no avail. Such an easy exercise once you know $U$ is the unit disc. Thanks. – PeterM Jan 23 '13 at 23:42

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