How to evaluate a zero of the Riemann zeta function? Here is a super naive question from a physicist:

Given the zeros of the Riemann zeta function,
  $\zeta(s) 
= \sum_{n=1}^\infty n^{-s}$,
  how do I actually evaluate them?

On this  web page I found a list of zeros. Well I guess if the values, call one ${a_\text{zero}}$, are just given in decimal expansion, then I'd have to run a program and see how it approaches zero 
$$\zeta({a_\text{zero}}) =
\sum_{n=1}^\infty n^{-{a_\text{zero}}} =
\frac{1}{1^{a_\text{zero}}} + \frac{1}{2^{a_\text{zero}}} + \frac{1}{3^{a_\text{zero}}} + \cdots\approx 0. \qquad ;\Re(a)>1$$
But are there also analytical solutions, let's call one ${b_\text{zero}}$, which I can plug in and see $\zeta({b_\text{zero}}) = 0$ exactly? Specifically the non-trivial ones with imaginary part are of interest. What I find curious is that these series, with each term being a real number multiplied by some $n^{-i\cdot \text{Im}(b_\text{zero})}$, then also kind of resemble Fourer series.

The Riemann Zeta function has zeros as follows:
(1) From the Euler Product follows $\zeta(s)\not=0$ for $\mathrm{Re}\,s>1$.
Taking the functional equation into account results that the only zeros outside the critical strip $\{ s\in\mathbb C\mid 0\leq\mathrm{Re}\,s\leq1\}$ are the trivial zeros $-2,-4,-6,\ldots$
(2) Beyond the trivial zeros the Zeta-Function has also zeros within the critical strip $S = \{ s \in \mathbb{C} \mid 1 > \mathrm{Re} \, s > 0 \} $. These are the non-trivial zeros. Little is yet known about these and your link to the numerical calculations of Andrew Odlyzko refer exactly to those zeros. There is not an analytical solution yet available.
To your point on the Fourier series analogy. It is possible to write out the Zeta function for $\mathrm{Re}\,s>1$ as a so called partition function that (resembles the form of a Fourier):
$$Z(T):= \sum_{n=1}^\infty \exp \left(\frac{-E(n)}{k_B T}\right) = \sum_{n=1}^\infty \exp \left(\frac{-E_0 \log n}{k_B T}\right) \equiv \sum_{n=1}^\infty \exp \left(-s \log n\right) = \sum_{n=1}^\infty \frac{1}{n^s} = \zeta (s)$$
and so forth see here>>>. But this was not the question I guess.
