Show that $f(z) = cg(z)$ is valid for two analytic functions $f$ and $g$ given a certain relationship between them Question: Suppose that functions $f$ and $g$ are analytic on the disc $A = \{z : |z| < 2\}$ and that neither $f(z)$ nor $g(z)$ is ever $0$ for $z \in A$.
$\frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$ for $n=1,2, ... ,$
show that there is a constant $c$ such that $f(z) = cg(z)$ for all $z \in A$.
Comments: This is one of the more theoretical problems from an old complex analysis exam. I am not sure which theorems I am supposed to use here. Am I supposed to take the integrals of both sides over the same curve and use the root-pole counting theorem? All input very much appreciated, so far everyone here has been of great help.
 A: Since $f$ and $g$ are holomorphic and never zero, $\frac{f'}{f}$ and $\frac{g'}{g}$ are again holomorphic. We know that $\frac{f'}{f} = (\log(f))'$ and so what we really have is that $(\log(f))' = (\log(g))'$ on the set $\{\frac{1}{n}: n\in\mathbb{N}\}$. There is a theorem that says that if two holomorphic functions agree on a set of points with an accumulation point, then they are equal. Therefore we have that $\log(f) = \log(g) + b$ for some $b\in\mathbb{C}$. Can you take it from here?
A: $\frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$ for $n=1,2, ... ,$ implies
$f'(\frac{1}{n})g(\frac{1}{n})-g'(\frac{1}{n})f(\frac{1}{n})=0$ for $n=1,2, ... ,$.
This implies that $\frac{f'(\frac{1}{n})g(\frac{1}{n})-g'(\frac{1}{n})f(\frac{1}{n})}{g(\frac{1}{n})^2}=\left(\frac{f(\frac{1}{n})}{g(\frac{1}{n})}\right)'=0$ and in turn that, since $f/g$, $(f/g)'$ are analytic and the latter agrees with the function $0$ on a bounded, convergent series of points, $(f/g)'=0$ on $A$, by the identity theorem and $f/g=c$ (on the set $A$) for some complex constant $c$.
