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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.

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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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why $0$ function? – Un Chien Andalou Apr 28 '13 at 9:54
@Tsotsi: It is unclear to me what your question means. Do you mean that you have trouble showing that $(f/g)'$ is the zero function, using the hypothesis in the question and the quotient rule? Or do you have trouble seeing how that implies that $f/g$ is constant? Or something else? – Jonas Meyer Apr 29 '13 at 0:43
why $(\frac{f}{g})'$ will be $0$ function? – Un Chien Andalou Apr 29 '13 at 5:39
@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. – Jonas Meyer Apr 29 '13 at 12:50

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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Here's what I have in mind. On an open disk $U$ centered at $a$, there are holomorphic $h$ and $k$ such that $\exp\circ h=f$ and $\exp\circ k = g$. The hypothesis and 2. implies that $h'=k'$ on $U$, so $h-k$ is constant on $U$, and therefore so is $\exp(h-k)=\frac{f}{g}$. By the identity theorem again, the result follows. – Jonas Meyer Apr 26 '11 at 0:29
Thanks for clarifying; I've deleted my first comment, because my confusions are now different. Local logarithms of $f$ and $g$ are among the local antiderivatives of $\frac{f'}{f}$ and $\frac{g'}{g}$, but I perhaps I am being pedantic. What I do not understand now is your second application of the identity theorem. If $D$ is not simply connected, there may be no holomorphic logarithms of $f$ and $g$ on $D$, so saying that $\log f = \log g +k$ on $D$ need not make sense. – Jonas Meyer Apr 26 '11 at 0:52
@Jonas: How about this. $log$ $f$ and $log$ $g$ are antiderivatives of $f'\over f$ and $g'\over g$ on some open set disk $U$ (which contains the point $a$). A disk is simply connected so the logarithm argument should hold and show that $f=cg$ on $U$. Then we extend $f=cg$ to all of $D$ by the Identity theorem. The difference here is that I don't claim anything about logarithms on all of $D$. – user3180 Apr 26 '11 at 1:16
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. – Jonas Meyer Apr 26 '11 at 1:26

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