Is $\zeta$ a conformal map on the right half-plane $\Re(s)>1$?

Question

The question is in the title: is the Riemann zeta function a conformal map on the right half plane where the Euler product is (proven to be) valid?

Answer 1

• As $\Re(s) \to \infty$, $\zeta(s) = 1+2^{-s}+O(3^{-s})$ thus it is not injective on $\Re(s) > \sigma $ for any $\sigma>1$. On strips $\Re(s) \in (\sigma_1,\sigma_2)$, $\zeta'$ is bounded and $\zeta(s+it)$ is almost periodic in $t$ thus it is again not injective.

  Moreover iff $|x| \in (\frac{1}{|\zeta(y)|}, |\zeta(y)|)$ then $\zeta(s)-x$ has infinitely many zeros on $\Re(s) \in (y-\epsilon,y+\epsilon)$, which follows from the Euler product and the $\Bbb{Q}$-linear independence of the $\log p$.

• $$\zeta'(s) = \sum_{n=1}^\infty (-\log n)n^{-s}=(-\log 2) 2^{-s}(1+\sum_{n=3}^\infty \frac{\log n}{\log 2} (n/2)^{-s})$$

  Let $a > 1$ such that $\frac{\zeta'(a)}{ (-\log 2) 2^{-a}} = 2$ then for $\Re(s) > a$, $$\sum_{n=3}^\infty |\frac{\log n}{\log 2} (n/2)^{-s}| < 1$$

  thus $\zeta'(s) \ne 0$ and $\zeta(s)$ is conformal (and locally biholomorphic) for $\Re(s) > a$.

  It is very plausible $\zeta'(s)$ has infinitely many zeros on $\Re(s) \in (a-\epsilon,a]$.

• For $ \Re(s) > a$ by absolute convergence we have the generalized Dirichlet series

$$\begin{eqnarray}\frac{1}{\zeta'(s)}& =& \frac{1}{(-\log 2) 2^{-s}}\frac{1}{1-(-\sum_{n=3}^\infty \frac{\log n}{\log 2} (n/2)^{-s})}\\ &=&\frac{1}{(-\log 2) 2^{-s}} \sum_{k=0}^\infty (-\sum_{n=3}^\infty \frac{\log n}{\log 2} (n/2)^{-s})^k \\ & = & \frac{1}{(-\log 2) 2^{-s}}(1+\sum_{k=1}^\infty (-1)^k \sum_{m =3^k}^\infty c_{k,m} (m/2^k)^{-s}) \\ &=& \sum_{r \in \Bbb{Z}[1/2]_{\ge 2}} b_r r^{-s} \end{eqnarray} \tag{1}$$