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Let $f$ and $g$ be two holomorphic functions in a connected open set $D$ of the plane, which have no zeros in $D$; if there is a sequence $(a_n)$ of points of $D$ such that

$\lim a_n = a, \qquad a \in D \quad\text{ and } a_n \neq a\text{ for all }n,$

and if \displaystyle\frac{f'(a_n)}{f(a_n)} = \frac{g'(a_n)}{g(a_n)}\qquad\text{ for all }n

show that there exists a constant $c$ such that $f(z) = c\cdot g(z)$ in $D$.

I've been penciling around with this question for a while and still can't get it. From what I can tell, the following theorems may be useful:

  1. If $f$ is holomorphic in $D$ then $f$ is analytic in $D$ (can be represented as a power series)

  2. If $\lim a_n = a$ for complex numbers $a_n$ and $a$ such that $a_n \neq a$ for all $n$, and $a$, $a_n$ all lie in $D$, then for an analytic function $f$ in $D$ and $f(a_n) = 0$ for all n, then $f = 0$ is the zero function.

2 Answers 2

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You can apply the hypothesis, the quotient rule, and 2. to show that \left(\frac{f}{g}\right)' is the zero function. This means that $\frac{f}{g}$ is constant, and the result follows.

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    @Tsotsi: Because the set of zeros of the analytic function $f'g-fg'$ has a limit point in the connected domain, it is identically zero, by "2" listed in the question.2013-04-29
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By the identity theorem, the functions f'(z)\over{f(z)} and g'(z)\over{g(z)} are equal on all of $D$. These functions are holomorphic (since the denominators have no roots in $D$) so they locally have an antiderivative at any point: $log$ $f(z)$ and $log$ $g(z)$ respectively. When two functions are equal their if antiderivatives differ by a constant so we have $log$ $f(z)$=$log$ $g(z)$ + $k$. Once again by the identity theorem, if $log$ $f(z)$=$log$ $g(z)$ + $k$ on an open set, then this also holds on all of $D$. It now follows that $f$ and $g$ differ by a multiplied constant.

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    Yes, I agree with that (and it is more or less the argument in my first comment). Of course, $\log f$ and $\log g$ are not uniquely determined on $U$, but you can choose some$log$of each and apply the argument.2011-04-26