Can sage find the poles of a complex function, and the residues at those poles?

edit

Apparently Maxima includes this, and we didn't even know it! See for instance here and the official docs.

(%i1) residue (s/(s**2+a**2), s, a*%i);
(%o1)                           1/2

In Sage, that would be doable.

sage: f(x)=1/x
sage: f.maxima_methods().residue(x,0)
1
f(x)=1/(x^2+1)
f.maxima_methods().residue(x,i)
-1/2*I

Well, we should wrap this!

This does not solve the problem of finding poles. Maple apparently has this, though perhaps not for such an example as the original one.

This won't work with a complicated example such as yours, but perhaps it will give you some ideas:

sage: f = sinh(x)/(sin(x)-1/2)
sage: (1/f == 0).solve(x)
[x == 1/6*pi]
sage: pole = _[0].rhs()
sage: (f*(x-pole)).limit(x=pole)
1/3*(e^(1/3*pi) - 1)*sqrt(3)*e^(-1/6*pi)
sage: f.taylor(x,pole,-1)*(x-pole)
2/3*sqrt(3)*sinh(1/6*pi)

the second line solves 1/f(x)=0, which for meromorphic functions is like solving f(x)=infinity. We then store the pole found in "pole". The last two lines calculates the residue, assuming the pole is simple (the two answers are identical).

Here is an example with a non-simple pole:

sage: f = sinh(x)/(sin(x)-1/2)-1/(x-pi/6)^2
sage: (1/f == 0).solve(x)
[x == 1/6*pi]
sage: pole = _[0].rhs()
sage: f(x=x+pole).taylor(x,0,-1).coeff(x,-1)
2/3*sqrt(3)*sinh(1/6*pi)

The substitution "x=x+pole" is needed - f.taylor(x,pole,-1).coeff(x-pole,-1), which should have given the same answer returns zero, since some automatic simplification takes place and messes things up:

sage: 1/(x-pole)                                  
-6/(pi - 6*x)

hope this helps!