when does $h$ exist?

Question

Let $\zeta(s)$ denote the Reimann zeta function in the critical strip. It is easy to see that $$ \zeta(s) = 0 \Longleftrightarrow \Re(\zeta(s))+\Im(\zeta(s)) = 0 ~~~~ \text{and} ~~~~~~ \Re(\zeta(s)) \Im(\zeta(s))=0$$.

Define $f: \mathbb{C}\rightarrow \mathbb{R}$ and $g:\mathbb{C}\rightarrow \mathbb{R}$ by the rules $$f(s) =\Re(\zeta(s))+\Im(\zeta(s)),~~~~~~~~~ g(s) = \Re(\zeta(s))\Im(\zeta(s)).$$

Is there a function $h$ so that $h(g(s)) = f(s)$?

I tried to start with partial sums of zeta function for finite value then take the product and after take the limit as so

$$\lim_{k\rightarrow \infty}{\biggl( \sum_{n=1}^{k} \frac{(-1)^{n-1}}{n^{\Re(s)}} \cos(\Im(s) \ln(n))\biggr)\biggl( \sum_{n=1}^{k} \frac{(-1)^{n-1}}{n^{\Re(s)}} \sin(\Im(s) \ln(n))\biggr)} = \Re(\eta(s))\Im(\eta(s))$$ with $\eta(s)$ the Dirichlet eta function.

maybe proceed by expanding this finite product and taking limit after but it does not seem to be going in a useful direction. I want to try and find for which values of $s$ does $h$ exist. If it does always exist try to find it.

Answer 1

There is no such function $h$, not just for the Riemann zeta function in the critical strip but for any non-constant holomorphic function $v:D\to\mathbb{C}$ defined on an open disk $D\subset\mathbb{C}$. Indeed, let $f=\Re(v)+\Im(v)$ and $g=\Re(v)\Im(v)$. Assume that a function $h:\mathbb{R}\to\mathbb{R}$ exists such that $h(g(s))=f(s)$ holds for all $s\in D$. Then $g(s)$ determines $f(s)$, hence it also determines $v(s)$ up to two choices. On the other hand, the range $\{v(s):s\in D\}$ is open, hence it contains whole arcs on certain hyperbolas $\{x+iy:(x,y)\in\mathbb{R}^2,\ xy=c\}$. Contradiction.