# Elliptic Functions, Residue Computation, Same zeros and poles of same orders

I am trying to understand the attached:

I know that if two functions have zeros and poles at the same point and of the same order then they differ only by a multiplicative constant, so that is fine, as both have a double zero at $$z=w_j/2$$ and a double pole at $$z=0$$.

But I don't understand at all the idea before determining what the constant $$C$$ should be? I thought that perhaps we had set the residues at the double pole $$z=0$$ equal, but this is given by:

$$\frac{1}{2}lim_{z \to 0} \frac{d}{dz}(z^2f(z))$$,

whereas it looks like we've compared

$$lim_{z \to 0} z^{2} f(z)$$,

so unless we have some reason to take the derivative outside the limit or something, I don't understand what we've done, and even whether my thoughts are on the right track and the residues are being compared?

Many thanks in advance.

Just bring the fraction of the RHS to the LHS such that only $C$ remains on the RHS. Multiply both the nominator and the denominator of the LHS by $z^2$ and take the limit $z\rightarrow 0$. Then you get $$C=\frac{1}{-1}=-1.$$
• Thanks. I guess what I'm trying to get at is what is the significance of taking the limit at this point $z=0$? So we are not comparing the residues just continuity at $z=0$, could we have chosen any point? Thanks – yourlazyphysicist Apr 4 '17 at 10:59
• ahh okay thanks, and $z=0$ is the easiest to work with here? – yourlazyphysicist Apr 5 '17 at 10:29
• It has to be multiplied by at least $z^2$ though because that is the order of the pole and any power smaller the limit would be undefined? – yourlazyphysicist Apr 7 '17 at 16:47