$\exists \enspace c \in \mathbb{C} : f(z) = cg(z)$

Question

$U = \{z \in \mathbb{C}: |z|<1\}$ and $\enspace f,g$ holomorphic functions on $U$ such that $f(z) \neq 0 \neq g(z)$. Moreover, it holds that $\forall n>1$

$\enspace \frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$

Now I have to show that: $\exists c \in \mathbb{C} \enspace : \forall z \in U \enspace \enspace f(z) = cg(z)$.

I tried to use the identity theorem, but I am not sure how to apply it. Can anyone give me a hint? Thanks!

Answer 1

By the identity theorem, $\frac{f'}f=\frac{g'}g$. But\begin{align}\frac{f'}f=\frac{g'}g&\iff f'g-fg'=0\\&\iff\left(\frac fg\right)'=0\\&\iff \frac fg\text{ is constant.}\end{align}

Answer 2

Observe that $$\left(\frac{f}{g}\right)' =0 $$ because $$\left(\frac{f}{g}\right)'\left(\frac{1}{n}\right) =0. $$