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

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I am trying to understand the attached:

enter image description here

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.

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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 $$C=\frac{1}{-1}=-1.$$