In light of the previous application, we might want to try to find such a formula for:

(z)(z0)(z)(z0)

The only possible singularities are at 0,z0,z0, since is even, and 0 is a pole of (z) mod L. We can see that z0 is removable by dividing the numerator and denominator by zz0 and taking the limit, or by noting that 1;;z0 is a zero of order 1 in the denominator, and a zero of order at least one in the numerator.

We can further calculate the residues of f at 0 and at z0, and we get 2 and 2 respectively. Applying the previous theorem, we get
f(z)=2ζ(z+z0)2ζ(z)2ζ(z0).

Powered by Forestry.md