# Showing $\exists~$ some $c$ such that $f(z)=cg(z)$ [duplicate]

Possible Duplicate:
Holomorphic functions and limits of a sequence

Hi There,

I was looking through an old text of mine, just refreshing myself on some material and I came across an interesting exercise statement that looks promising to understand. It just have been some time and I do not know if I am/even know how to approach the problem correctly. Some guidance of a solution for my thoughts to think on would be appreciated. The question is as stated:

Suppose that $f$ and $g$ are analytic on the disk $A=\{z \text{ such that }|z| \lt 2 \}$ and

that neither $f(z)$ nor $g(z)$ is ever $0$ for $z \in A.$ If

$$\frac{f^{\;'}(1/n)}{f(1/n)}=\frac{g'(1/n)}{g(1/n)} \quad {\text{ for }} \quad n=1,2,3,4 \ldots,$$

could it be shown that there is a constant $c$ such that $f(z)=cg(z)~~\forall ~~ z \in A.$

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## marked as duplicate by Jonas Meyer, t.b., Chandru, night owl, AryabhataJun 2 '11 at 8:14

@Jonas Meyer: Thanks, okay opportunity for another questions I had wanted to ask for awhile. – night owl Jun 2 '11 at 7:39

Denote $h(x)=\frac{f^{'}(x)}{f(x)}-\frac{g^{'}(x)}{g(x)}$, which is holomorphic in $A$, since $f,g$ are never zero and analytic on $A$. The statement says that $h(\frac{1}{n})=0,\ \forall n \geq 1$. There is a theorem in complex analysis:

Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically $0$.

Since this is the case for $h$, we have $h(x)=0,\ \forall x \in A$. This means that $f^{'}(x)g(x)-f(x)g^{'}(x)=0,\ \forall x \in A$, and therefore $(f/g)^\prime(x)=0,\forall x \in A$. Therefore $f/g=c$.

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Yes, this is simpler than my argument. – Andrés E. Caicedo Jun 2 '11 at 7:43
+1..For simplicity. I have to search for that theorem as it seems I lost sight of it, i.e., erased from my internal hard drive :) – night owl Jun 2 '11 at 7:47
@night owl: See the Wikipedia page on the identity theorem for holomorphic functions. @Beni: very nice answer! – t.b. Jun 2 '11 at 7:49
Thank you. It seemed the most natural :) – Beni Bogosel Jun 2 '11 at 7:59
@night owl: I don't really know what you mean. I have quite a good memory and thus remember what to look for using Google. Just add site:math.stackexchange.com to your search terms in order to restrict the search to this site. The built-in search here is not very useful to me (it searches using OR instead of AND). This time I remembered seeing that question but Jonas spotted the duplicate as he was one of the answerers :) – t.b. Jun 2 '11 at 8:17

If $f$ is analytic in a disk but never 0, it is $e^h$ for some analytic $h$. This is not a trivial statement. (But your book probably spent a good deal of time getting here by the time the exercise is stated.) I expect this should suffice as a hint. If you want to work out the solution on your own, stop reading here. The rest of the argument is below.

Note that if $f=e^h$ then $f'=e^h h'=fh'$, or $h'=f'/f$. You are told that $g$ also behaves in the same way, so $g'/g=j'$ for some analytic $j$ (with $e^j=g$).

Then $h'(1/n)=j'(1/n)$ for all $n$ and, by continuity, $h'(0)=j'(0)$. But then $h'=j'$. (Because they coincide in a set that has a limit point inside their domain. See a proof here.)

So $h(z)=j(z)+k$ for a constant $k$ (again, I assume by this point you know that analytic functions on a disk with derivative equal to 0 are constant). But then $f=e^h=e^k e^j=cg$ where $c$ is the constant $e^k$.

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Nice, yes the preliminary goes through this as you mentioned, just couldn't quite wrapped it all together for the piece: $(f/g)'(1/n)$ – night owl Jun 2 '11 at 7:53