4 inserted a factor 4 in the exponent of T

1) Yes, this is still possible

2) No. We know that they have to distributed with low density, i.e. the number of zeros $z$ with $\Im z < T$ with $\Re z > \sigma >1/2$ is bounded by $T^{\sigma(1-\sigma) T^{4\sigma(1-\sigma) + \epsilon}$ for $\epsilon >0$. So no uniform distribution is possible, since the gaps between consecutive zeros has to grow to infinity.

For sharper results in this direction, I suggest the first chapter of Joern Steuding "Universality of L-functions". This book actually explains pretty good what is going on in the critical stipe off the line $\Re s = 1/2$.

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1) Yes, this is not knownstill possible

2) No. What is We know that they have to distributed with low density. E.g, i.e. the number of zeros $z$ with $\Im z < T$ with $\Re z > \sigma >1/2$ off the critical line does not grow like $T$, but is bounded by $T^{\sigma(1-\sigma)}$ if I recall correctly. T^{\sigma(1-\sigma) + \epsilon}$for$\epsilon >0$. So no linear uniform distribution is possible, since the gaps between consecutive zeros has to grow to infinity. For more sharper results on in this direction, I suggest the first chapter of Joern Steuding "Universality of L-functions". This book actually explains pretty good what is going on in the critical stipe off the line$\Re s = 1/2$. 2 added 63 characters in body This is not known. What is know that they have to distributed with low density. E.g. the number of zeros with$\Im z < T$with$\Re z > \sigma >1/2$off the critical line does noit not grow like$T$. T$, but is bounded by $T^{\sigma(1-\sigma)}$ if I recall correctly. So no linear uniform distribution is possible. For more results on this, I suggest the first chapter of Joern Steuding "Universality of L-functions".

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