Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Holomorphic functions and limits of a sequence

Let $\Omega$ a domain and $f,g$ holomorphic function in $\Omega$. Suposse that $\exists$ a sequence $\{a_n\}$ in $\Omega$ convergent to a point $a$ of $\Omega$: $$a_n\neq a, \forall n, \\\\\\f'(a_n)g(a_n)=g'(a_n)f(a_n), \forall n, $$ with $g(a)\neq0$.

Probe that $f$, $g$ are linearly dependent.

share|cite|improve this question

marked as duplicate by Jonas Meyer, Did, Austin Mohr, Potato, Pavel M Jan 10 '13 at 3:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You have two holomorphic functions agreeing on a set with a limit point. Therefore they are equal. So $f'g = g'f$. Hence $$(f/g)' = {g'f - f'g\over g^2} = 0$$ Therefore $f/g$ is constant.

share|cite|improve this answer

If $g(a)\neq0$, then $g(a_n)\neq 0$ for $n$ sufficiently large. Therefore we can consider the ratio $h=f/g$; this function is holomorphic in a neighborhood of $a$ and $h'=(f'g-g'f)/g^2$. By hypothesis, $h'(a_n)=0$, for $n$ sufficiently large, hence, as the limit point of $a_n$ is in $\Omega$, you know that $h'\equiv0$ in $\Omega$, therefore, in a small disc centered in $a$ you can find $c\in\mathbb{C}$ such that $h\equiv c$ on that disc. Therefore, in a small neighborhood of $a$, $f=cg$, i.e. $f-cg\equiv0$ in that neighborhood, therefore $f-cg=0$ on the connected component of $\Omega$ containing $a$; as domains are usually considered connected, we have that $f=cg$ on $\Omega$.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.