I mean the non-periodic "oscillations" (besides slow calculation) for small y, that would give no chance for a purely numerical routine: one would need estimates / bounds. And think one has to invest some theory (where I do not even see, why the integral exists). I give up.

I have some doubts that a direct way has a chance if looking at the following for y0 = 0 and y0 = 1000 (but I give up to look for reformulations, missing needed theory):

#Digits:=15;
#y0:=1000;
abs(Zeta(x+I*y));
plot3d(%, y=y0..y0+100, x=2 .. 6, axes=boxed);

@eithne you may also consider to add some more informations about processing (which and where data are stored or transmitted to 3rd parties - not that I personally use the service)

For me it also works (Win 7 64 bit, Maple 18.00 32 bit, classic)

I like that (is it due to GMP as library?)

@Markiyan Hirnyk You are right, I answered too hastily, sorry

Yes. But the whole system? I do not see, why it has no components of higher dimensions. May be I did not get it correctly: alpha=nu/(1+nu), lambda=0, mu=1-nu defines a curve - gives it all solution?

I played a bit more: it is linear * quadratic * cubic and Maple needs more time than I am patient to 'proof' the solution has dim = 0 (over the Complex).

Restricting to one of those degrees the solution is a surface (while all variables occure for deg = 2,3)

One idea might be to 'diagonalize' the quadratics, where one needs to pray that one can do it simulanously.

May be it would be better to present the context of the problem.

PS: I would prefer English for discussions, even if it may be broken like mine, just as a kind of Esperanto.

Using 'factor' one can see, that those are products of maximal degree = 3, each variable and up to combinatorical care thus can be reduced even a bit more, similar as you suggested. Have not done it.

You want ordering by 2,3,1, i.e, plots:-display( all_plots[2], all_plots[3], all_plots[1] )

Not sure, but for extra options (which one?) the prior ones 'dominate', try with 'title' for a test

ok (found it, 10 years old)

I wonder where you got this lovely and cynical task from :-)

Mostly I use 'evala', which works as well in this case,

convert(U,elementary); evala(%);

gives the same

a sheet for that, with some tweaks

zero_of_a_tan(z)-tanh(z).mws

zero of a tan(z)-tanh(z).mws

The zeroes on the real and imaginary axes are the 'same':

  'eval(tan(y)-tanh(y), y=I*x)' = -I*(tan(x)-tanh(x)); 
  is(%);

                                 true

To carry over Carl's argument to the complex one should use periodics
again and then has to prove, that there is exactly 1 zero, now in the
rectangle = Pi/2 * (-1-I) ... Pi/2 * (+1+I) Pi * (-1-I) ... Pi * (+1+I).

For this re-write as cos and sin, take the numerator of the according
expression (a rational function in trigonometrics),

  f:= z -> -sinh(z)*cos(z)+sin(z)*cosh(z);

This is holomorphic (and entire), series(f(z),z=0) shows that it has
a zero of order 3 in z=0.

http://en.wikipedia.org/wiki/Argument_principle shows the zero counting
integral and one has to show it equals 3 for domain = above rectangle.

It is somewhat tedious to write down the natural path for the boundary,
parameter t = 0 .. 1

Let be H the integrand for the zero counting integral. Maple can not
solve the task symbolically (and I think it makes an error if using
a circle instead), and needs some help to do it numerically:

For that plot Re(H) and Im(H) to see periodics of 1/4 w.r.t., which
can be confiremd symbolically.

So one needs only the integrals over t = 0 .. 1/4 (now 4* those) for
the real and imaginary parts)

Choosing a relativ error now Maple succeeds and confirms the assertion.