Is there an explicit expression for the imaginary part of some nontrivial zero of zeta, in terms of wellknown constants, such as say $\gamma$ or $\pi$ say ?

Write $\rho = \frac12 + i \gamma$ for a nontrivial zero of a primitive Lfunction. ("Primitive" means that it can't be written as the product of other Lfunctions.) It is generally believed that: a) If $\gamma\not=0$ then $\gamma$ is transcendental. b) If $\gamma\not=0$ then $\gamma$ is algebraically independent of every wellknown constant and every other zero of every primitive Lfunction (except when the Lfunction has real coefficients, in which case $\frac12  i \gamma$ is also a zero). As far as I know, nobody has any clue how to prove these conjectures. Clarification added later: what definition of Lfunction are we using? Greg Martin's comment (below) refers to $L(s+ i y)$ where $L(s)$ is an Lfunction and $y$ is real. While it is true that for some definitions of "Lfunction" the set of Lfunctions is closed under that operation, that is not what I intended. For the Lfunctions in my answer above, the Euler product axiom can be written as: There is a Dirichlet character $\chi$, the "central character" of the Lfunction, such that \begin{equation} L(s)= \prod_{p \, {\rm prime}} F_p(p^{s})^{1}, \end{equation} where $F_p$ is a polynomial of the form \begin{equation} F_p(z)=1a_p z + \cdots + (1)^d\chi(p) z^d . \end{equation} Here $d$ is the degree of the Lfunction. Note that I have normalized the Lfunction so that the functional equation relates $s$ to $1s$. All known Lfunctions satisfy that axiom, and this formulation tells you how to select the distinguished member of the family $L(s+i y)$. 


There is no explicit value for the imaginary part of the nth zero. However it satisfies a simple transcendental equation for each n, whose solution is well approximated by the Lambert function. See LeClair and Franca on arXiv, math.NT 

