**Edit** Major rewrite since Johan Andersson observed the original
question is trivial because of vanishing of coefficients.

From this question

$$ \zeta(1-x) = \frac{2(\zeta'(x)\Gamma(x/2)+\Gamma((1-x)/2) \zeta'(1-x)\pi^{x-1/2}) )}{\Gamma((1-x)/2) \pi^{-1/2+x}(2\log\pi -\psi((1-x)/2)-\psi(x/2))} \qquad (1)$$

On the critical line $\zeta'(1/2 - i t) = |\zeta'(1/2+ i t)|^2 / \zeta'(1/2+ i t)$ so $\zeta'(1/2 - i t)$ can be eliminated, leaving on $\zeta(1/2+i t),\zeta'(1/2+i t)$ and modulus of the derivative,leading to

$$ \zeta \left( 1/2+it \right) = 2\,{\frac {|\zeta'\left(1/2+it \right)|^2/\zeta'\left(1/2+it \right) \Gamma \left( 1/4-1/2\,it \right) +\Gamma \left( 1/4+1/2\,i t \right) \zeta' \left(1/2+it \right) {\pi }^{-it}}{\Gamma \left( 1/4+1/2\,it \right) {\pi }^{-it} \left( 2\,\ln \left( \pi \right) - \psi \left( 1/4+1/2\,it \right) -\psi \left( 1/4-1/2\,it \right) \right) }} \qquad (D) $$

So on the critical line $\zeta(1/2+it)=f(\zeta'(1/2+it))$ though the complex modulus complicates things.

(Fishing expedition) Can (D) be solved for zeta, unlikely giving alternate expression for zeta on the critical line?

sage/mpmath code in case of typos

```
def difide1(t):
"""
differential equation via |zeta'(s)| on the critical line
should return zero for real $t$
"""
from mpmath import gamma,zeta,log,psi
J=mpmath.j
Pi=mpmath.pi
t=mpmath.mpc(t)
return 2*( ( mpmath.fabs(zeta(1/2+J*t,derivative=1))**2/zeta(1/2+J*t,derivative=1) )*gamma(1/4-J*t/2)+gamma(1/4+J*t/2)*zeta(1/2+J*t,derivative=1)*Pi**(-J*t))/(gamma(1/4+J*t/2)*Pi**(-J*t)*(2*log(Pi)-psi(0,1/4+J*t/2)-psi(0,1/4-J*t/2 ) )) - zeta(1/2+J*t)
```