Is there an explicit expression for the imaginary part of some non-trivial zero of zeta, Is there an explicit expression for the imaginary part of some non-trivial zero of zeta, in terms of well-known constants, such as say $\gamma$ or $\pi$ say ?
 A: Write $\rho = \frac12 + i \gamma$ for a nontrivial zero of a primitive L-function.
("Primitive" means that it can't be written as the product of other L-functions.)
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 well-known
constant and every other zero of every primitive L-function (except when the
L-function 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 L-function are we using?
Greg Martin's comment (below) refers to $L(s+ i y)$ where $L(s)$ is an L-function
and $y$ is real.  While it is true that for some definitions of "L-function" the
set of L-functions is closed under that operation, that is not what I intended.
For the L-functions in my answer above, the Euler product axiom can be written as:

There is a Dirichlet character $\chi$, the "central character" of the 
L-function, 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)=1-a_p z + \cdots + (-1)^d\chi(p) z^d .
\end{equation}

Here $d$ is the degree of the L-function.  Note that I have normalized the L-function
so that the functional equation relates $s$ to $1-s$.
All known L-functions satisfy that axiom, and this formulation tells you how to select
the distinguished member of the family $L(s+i y)$.
A: There is no explicit value for the imaginary part of the n-th 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
