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If $K$ is a number field, whose Galois closure over the rationals has degree 24 or so, and whose discriminant is around $163^4$, then what is a numerically efficient way of computing the first few zeros of its zeta function on the critical line?

I tried in pari but pari seems to choke on zetakinit.

I tried in magma and got much further. I can create the number field, and use the LSeries command to compute some form of the $L$-function. I can now evaluate the $L$-function at pretty much any point I want on the critical line, and use things like LSetPrecision to warn magma that I'm going up the critical line. I have no feeling for these things though; I don't even know how far I might expect to look up the line for the first, say, five zeros. The main problem I have though is that I'm just naively evaluating the function at some random points, and each evaluation might take a minute, and I evaluate the function at a point and it's non-zero and now I don't even know whether to move up or down.

Are there any other computer algebra packages that might be able to help me out?

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(naïve question) Would the $L$-function you have happen to admit a decomposition a bit like decomposing Riemann $\zeta$ in terms of Riemann-Siegel functions, or are they too complicated for this approach? – J. M. Apr 30 '11 at 0:21
@J.M.: I don't know any more about Riemann-Siegel functions than is mentioned in the link, but it seems to me that the Riemann-Siegel function associated to the classical zeta function is just $Z(t)=\pm|\zeta(1/2+it)|$, and this notion could perhaps be generalised to any holomorphic function at all, zeta or not. – Kevin Buzzard Apr 30 '11 at 8:26
I remember asking (here and on MO) about the possibility of such decompositions; I haven't fully digested the answer given by Stopple @ MO, but it seems the things he's talking about might be of use to you as well... – J. M. Apr 30 '11 at 8:30
This question has been answered at… – Kevin Buzzard May 1 '11 at 8:02

As Kevin Buzzard noted in May, this question has been answered at

So this is no more an open question.

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