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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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See here for a nice outline: – t.b. Jun 1 '12 at 11:16
I've read that and don't really understand it at all! I think that this question is meant to be solved much more simply, since we haven't really talked about any of the theory in the course. – Edward Hughes Jun 1 '12 at 11:18
You could simply use the expansions and verify that the principal part vanishes, then apply Liouville (note that $f$ is $\Lambda$-periodic by definition). This makes a straightforward if somewhat messy calculation. – t.b. Jun 1 '12 at 11:20
Maybe I'm being stupid but I don't see how that tells me how many zeroes I have! – Edward Hughes Jun 1 '12 at 11:21
Okay have found two more, namely $z=-2w$, $z=-\frac{1}{2}w$ but these put different constraints on $w$ to the ones I've been given! – Edward Hughes Jun 1 '12 at 11:22

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