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Consider the function

$ \det\left( \begin{array}{ccccc} &1 &\wp(z) &\wp'(z) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(-z-w) &\wp'(-z-w) \end{array} \right)=f(z) $

I'm trying to prove that it has at least $6$ distinct zeroes if $w\notin\frac{1}{3}\Lambda$ and at least $5$ zeroes with multiplicity otherwise. Then by a degree argument $f$ is obviously identically zero. I've got no idea how to show this though! Obviously $f$ has a zero at $z=w$. Apart from that I know the zeroes of $\wp'$ but they seem to be of no help. It seems like I'm meant to try $z=3w$ but this doesn't work! A hint would be greatly appreciated!

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    Yeah I see, thank you :). I agree that is probably a better way of showing it, but it's not really how I need to prove it! (I've put up a clarifying comment to that effect)2012-06-01

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