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

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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##
 
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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}?
 
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anuttarasammyak said:
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
 
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?
 
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