Prove or Disprove: A holomorphic function with the given property exists

Question

Does there exist a holomorphic function $f:U_{1+\epsilon}(0)\rightarrow \mathbb{C}$ for all $n\in \mathbb{N}_{\ge 2}$ such that

$$ f\left(\log\left(1+\frac{1}{n}\right)\right) = \left(\frac{1}{n^2}-1\right)\left(1+\frac{1}{n}\right). $$

where $U_{1+\epsilon}(0)$ is the open disk of radius $1+\epsilon$ centered at $0$.

I tried to use the identity theorem for holomorphic functions, but don't know how to define a function $g$ such that $g(\log(1+\frac{1}{n})= \left(\frac{1}{n^2}-1\right)\left(1+\frac{1}{n}\right)$.

Answer 1

Let's construct the required function "by hand". Let $z = \log \left( 1 + \frac 1 n \right)$. Then $\frac 1 n = \Bbb e ^z - 1$, so that

$$f(z) = \Big( (\Bbb e ^z - 1)^2 - 1 \Big) \Big( 1 + (\Bbb e ^z - 1) \Big) = \Bbb e ^{2z} (\Bbb e ^z - 2)$$

which is clearly seen to be holomorphic (in fact, it is entire - i.e. holomorphic on the whole $\Bbb C$, not just on $U_{1 + \epsilon}$.

Answer 2

Hint: The identity theorem suggests $f(\log(1+z)) = (z^2-1)(1+z)$ will hold. Assume it does, find an inverse to $\log(1+z),$ and see what happens.