How can I take double ResRes at a singularity in the residue theorem?

[mrsaturn]

Homework Statement

Let $g(z_1,z_2)$ be a rational function where the only possible singularities are at $z_1 = 0$, $z_2 = 0$, and $z_1 = z_2$. Verify the following: $\text{Res}_{z_2 = 0}\text{Res}_{z_1 = z_2} g(z_1,z_2) =- \text{Res}_{z_1 = 0}\text{Res}_{z_2 = z_1} g(z_1,z_2)$

Relevant Equations

$\text{Res}_{z_2 = 0}\text{Res}_{z_1 = z_2} g(z_1,z_2) =- \text{Res}_{z_1 = 0}\text{Res}_{z_2 = z_1} g(z_1,z_2)$

I think we should be able to verify this by the residue theorem, but I'm having trouble applying it to the case when there is a singularity at $z_1 = z_2$

 

[anuttarasammyak]

To understand the problem exactly you mean for an example
$$g=\frac{1}{z_1}-\frac{1}{z_2}+\frac{1}{z_1-z_2}$$?

 

[mrsaturn]

To understand the problem exactly you mean for an example
$$g=\frac{1}{z_1}-\frac{1}{z_2}+\frac{1}{z_1-z_2}$$?

Yeah exactly

 

[anuttarasammyak]

Thanks. Then I would like to know how shall I take double ResRes as you do in your formula. Could you show me how to do in this example?