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It is hard to understand what you have and what you want. Here is a simple piece of code getting the residue for that $W_{1,1}$, in code W11. Since i hate (typing) $\sqrt{2a}$ i will use $b$ instead.

var('z,b,z1');
assume(b > 0);

y = lambda z: 2*arcsinh( z / b ) / sqrt(z^2 + b^2)
K = lambda z: 1 / z / (y(z) - y(-z))
W02 = lambda z1, z2: 1/(z1 - z2)^2
W11 = lambda z1: ( K(z) / (z - z1) * W02(z, -z) ).residue(z == 0).canonicalize_radical()

print(f'W11(z1) = {W11(z1)}')

This prints:

W11(z1) = -1/48*(3*b^2 + 2*z1^2)/z1^4

There is nothing new involved while taking residues when implementing all the needed recursion. (Please provide code in a follow-up question, if something does not work on the path. Please try to reduce the questions to one issue each, and best this issue is of programatical nature.)

It is hard to understand in one breath what you have tried and what you want. finally want.

Here is a simple piece of code getting the residue for that $W_{1,1}$, in code W11. The further implementation has from the point of view of the programmer nothing new to show up. Since i hate (typing) $\sqrt{2a}$ i will use $b$ instead.

var('z,b,z1');
assume(b > 0);

y = lambda z: 2*arcsinh( z / b ) / sqrt(z^2 + b^2)
K = lambda z: 1 / z / (y(z) - y(-z))
W02 = lambda z1, z2: 1/(z1 - z2)^2
W11 = lambda z1: ( K(z) / (z - z1) * W02(z, -z) ).residue(z == 0).canonicalize_radical()

print(f'W11(z1) = {W11(z1)}')

This prints:

W11(z1) = -1/48*(3*b^2 + 2*z1^2)/z1^4

There is nothing new involved while taking residues when implementing all the needed recursion. (Please provide code in a follow-up question, if something does not work on the path. Please try to reduce the questions to one issue each, and best this issue is of programatical nature.)