Specifically I am interested in the quotients $$\frac{\zeta'(\rho)}{\zeta'(1\rho)}=2(2\pi)^{\rho1}\Gamma(1\rho)\sin(\pi\rho/2).$$ Obviously they are in $\mathbb{T}$ for all known nontrivial zeros. But how often are these number $\pm 1$? I would find some tables of the derivative at the known zeros rather usefull, or even perhaps tables of the quotients above? I would be very grateful if somebody can provide me with a good reference. Thanks!

3$\begingroup$ I imagine you might be stuck calculating this yourself, which shouldn't be too hard (depending on your programming ability). In case you don't have it, here's a link to a table of zeros: dtc.umn.edu/~odlyzko/zeta_tables $\endgroup$ – B R Nov 11 '11 at 22:59

5$\begingroup$ Mathematica has about $10^7$ zeros built in as the function ZetaZero[n]. $\endgroup$ – Stopple Nov 11 '11 at 23:11

3$\begingroup$ I'd guess the real and imaginary parts are transcendental. After calculating the first 100 examples, none seem particularly close to $\pm1$ $\endgroup$ – Stopple Nov 11 '11 at 23:29

$\begingroup$ Thank you, Stopple. I haven't done the calculations because I don't have Mathematica on my laptop at the moment. I had guessed that the imaginary parts would get small quite quickly. What makes you guess the parts would be transcendental? $\endgroup$ – Kevin Smith Nov 12 '11 at 1:06

$\begingroup$ Interesting topic. I would like to know whether the real part of the first derivative of the Zeta function at the non trivial zeros of Zeta is stricly positive and if so, is there a proof for it. Thanks $\endgroup$ – HASSINE SAIDANE Dec 2 '13 at 14:25
To elaborate a little more, here's some Mathematica code:
ListPlot[Table[{rho = ZetaZero[n];z = N[2 (2 Pi)^(rho  1) Gamma[1  rho] Sin[Pi rho/2]]; {Re[z],
Im[z]}}, {n, 1, 100}], AspectRatio > Automatic]
Here's the output:

$\begingroup$ I am trying to get at whether these numbers tend to $1$. $\endgroup$ – Kevin Smith Nov 12 '11 at 1:09

1$\begingroup$ Your plot seems to suggest that they are repelled from $+1$, is there any further evidence of this conjecture? $\endgroup$ – Kevin Smith Nov 12 '11 at 1:19

3$\begingroup$ No, this phenomenon does not seem to appear looking at the first 1000. I'd guess they are dense on the circle. The imaginary parts of the zeros are conjectured transcendental  why would you conjecture anything else for these numbers? $\endgroup$ – Stopple Nov 12 '11 at 1:41

$\begingroup$ I am not conjecturing that the imaginary parts of the zeros are not transcendental (unless I am missing something implicitly). The quotient of the derivatives describes how much faster (or not, as you suggest) the imaginary part of $\zeta(1/2+it)$ goes to zero than the real part. Since the real part tends to be positive, I wondered if this phenomenon is is evident asymptotically as a tendency for the quotient to be near to $1$. Your response has been very helpful so I accept your answer... $\endgroup$ – Kevin Smith Nov 12 '11 at 1:58

$\begingroup$ but please do let me know if you find anything importantly, it wont be evident in plotting all the first 100, 1000, say, it would appear in looking at intervals starting further up the line. $\endgroup$ – Kevin Smith Nov 12 '11 at 2:01
Here is an answer in a few parts.
Ghaith Hiary has computed fairly large tables of zeros at large height, which are at
http://sage.math.washington.edu/home/hiaryg/page/index.html
I believe that each table has 10 million zeros starting at t = 10^n, for n from 12 to 28. There are tables of derivatives of Z'(t) at the zeros, but not zeta'(1/2 + it). At at zero, it is not hard to go from one to the other, though.
To compute what you want to compute, I think the formula you wrote down is not going to be computationally practical due to the exponential decay of the gamma function and the exponential growth of the sine function. (I think may already be precision issues in the picture in Stopple's answer.) If I calculate correctly (I hope I'm not embarrassing myself), then $$ \frac{\zeta'(1/2 + i\gamma)}{\zeta'(1/2  i\gamma)} = e^{2 i \theta(\gamma)} $$ where $\theta$ is the RiemannSiegel theta function. This is probably a better way to compute the quotient.
To make the above picture using sage, if you have the optional package database_odlyzko_zeta installed, you can use:
import mpmath mpmath.mp.prec = 200 L = [] for gamma in zeta_zeros()[:100]: z = exp(2 * CC.0 * CC(mpmath.siegeltheta(gamma))) L.append((z.real(), z.imag())) P = points(L) P.save('plot.png', figsize=[5,5])
Something similar will work if you want to parse Hiary's files.
As for whether there is any tendency for these numbers to be close to 1: I think there is not. This looks like the question of whether there is any asymptotic relation between the zeros of the zeta function and gram points, and I think that the answer is expected to be no, but I don't know if that is a theorem. (Those thoughts should not stop you from checking, though.)

$\begingroup$ Jonathan, thank you, this is very helpful. For starters, I must point out that my original hypothesis should have been $+1$, not $1$ (one of those silly latenight errors!). In any case, that there is no such tendancy ($\pm 1$) is equivalent to the statement that $u(t)\notin o(v(t))$ and $v(t)\notin o(u(t))$, where $u,v$ are the real and imaginary parts of derivative. In other words, $\liminfu/v(\rho)>0$ and $\limsupu/v(\rho)<\infty$, for $\beta=1/2$. I suppose the next question is whether Stopple's guess: that the set is dense in $\mathbb{T}$ is correct, which, as you pointed out... $\endgroup$ – Kevin Smith Nov 13 '11 at 10:51

$\begingroup$ amounts to knowing if $\theta(\gamma)$ is uniformly distributed in $[\pi,\pi]$. $\endgroup$ – Kevin Smith Nov 13 '11 at 10:52

$\begingroup$ Typo: replace $t$ with $\rho$ in the comment above. $\endgroup$ – Kevin Smith Nov 13 '11 at 11:02

$\begingroup$ In fact, uniform distribution would be much stronger, though still a possibility. I will reformulate this and post another question. $\endgroup$ – Kevin Smith Nov 13 '11 at 14:22

$\begingroup$ The link is unfortunately broken. Perhaps it should be people.math.osu.edu/hiary.1? $\endgroup$ – jeq Nov 1 '17 at 15:59