The Lehmers' zeros for $\zeta(\frac{1}{2} + it)$ are quite close to each other in the value of the ordinate $t$. I am aware of the constraint on the Hardy Z function for which a local positive minimum or a local negative maximum would disprove RH.
My question is does a zeta zero ever occur when either $Re(\zeta(\frac{1}{2} + it))$ plotted vs. $t$ is at a minimum or when the $Im(\zeta(\frac{1}{2} + it))$ plotted vs. $t$ is at a maximum?
Thanks
Note that if $\zeta(\frac{1}{2} + it)=u(t)+iv(t)$ and with $e^{i\theta(t)}\zeta(\frac{1}{2} + it)=Z(t)$ one has that $Z=u\cos \theta-v\sin \theta$ while $u\sin \theta+v \cos\theta=0$ so one gets $-v\sin \theta=u \sin^2\theta/\cos \theta=u/\cos \theta-u\cos \theta$ so $Z=u/\cos \theta$
But now if $Z(w)=0$ hence $u(w)=v(w)=0$ and $w$ is a minimum for $u$ so $u'(w)=0$ it follows that $Z'(w)=0$ and that is considered highly unlikely