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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: http://math.stackexchange.com/q/121662/53632012-06-01
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    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.2012-06-01
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    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.2012-06-01
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    Maybe I'm being stupid but I don't see how that tells me how many zeroes I have!2012-06-01
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    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!2012-06-01
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    I was simply suggesting alternative approaches. That's all...2012-06-01
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    Just to clarify the point of this question... I'm trying to prove that there are these zeroes, without resorting to proving that $f$ is identically zero first!2012-06-01
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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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