Holomorphic functions on unit disc Let $f,g$ be holomorphic on $\mathbb{D}:=\lbrace z\in\mathbb{C}:|z|<1\rbrace$, $f\neq0,g\neq0$, such that $$\frac{f^{\prime}}{f}(\frac{1}{n})=\frac{g^{\prime}}{g}(\frac{1}{n}) $$ for all natural $n\geq1$. Does it imply that $f=Cg$, where $C$ is some constant?
Let $A:=\lbrace\frac{1}{n}:n\geq1\rbrace$ and $h:=\frac{f^{\prime}}{f}-\frac{g^{\prime}}{g}$. Now, $h$ is holomprphic on $\mathbb{D}$ and disappears on a subset of $\mathbb{D}$ which has a limit point. Thus $h=0$, so $\frac{f^{\prime}}{f}=\frac{g^{\prime}}{g}$ on $\mathbb{D}$. 
Could someone help with the next steps? Or maybe $f$ doesn't have to be in the form described above?
 A: Notice that your last statement is equivalent to ${f'g-g'f}=0$, since $f,g \neq 0$. Now there is no harm in dividing that expression by $g^2$. You get ${f'g-g'f}{g^2}=0$. So, $(f/g)'=0$ and so your result follows.
A: If $f$ and $g$ are allowed to be $0$ somewhere (but not everywhere), the result is  still true.  Suppose $g(z) = z^k G(z)$ where $k$ is a nonnegative integer and $G$ is holomorphic on ${\mathbb D}$ with $G(0) \ne 0$.  Then $$\dfrac{g'(z)}{g(z)} = \dfrac{k}{z} + \dfrac{G'(z)}{G(z)} = \dfrac{k}{z} + O(1) \text{ as } z \to 0 $$
So from the sequence $(g'/g)(1/n)$ we can determine $k$.  Thus if $(g'/g)(1/n) = (f'/f)(1/n)$ for all $n$ we also have $f(z) = z^k F(z)$ (for the same $k$) where $F$ is 
holomorphic on ${\mathbb D}$ with $F(0) \ne 0$, and $F'/F = G'/G$.  Since you already 
proved the result for functions nonzero at $0$, it follows that $f/g = F/G$ is constant.
A: Hint: Since $$\frac{f'}{f}=\frac{g'}{g}$$ on $\Bbb D$, then $f'g-fg'=0$ on $\Bbb D$, and so $$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}=0$$ on $\Bbb D$. Can you get the rest of the way from there and see why $\frac{f}{g}$ is holomorphic on $\Bbb D$?
