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I've been really struggling with how to do this question. I can expand it by partial fractions but then have no idea what to do next.

Find the principal part of the innermost laurent expansion for: $ \frac{1}{z^2\sin(z)} $ about the point a = 0

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    I assume it means the expansion in an annulus r<|z| with $r$ minimal. (In this case it would be the annulus 0<|z|<\pi.2012-10-24

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How did you expand this into partial fractions? It is not a rational function. I would start with the power series expansion $\frac{1}{z^2 \sin z} = \frac{1}{z^2(z-z^3/6 \pm \ldots)} = \frac1{z^3}\frac{1}{1-(z^2/6\mp\ldots)}.$ Then use the geometric series expansion $\frac{1}{1-q} = 1+q+q^2+\ldots$ with $q=z^2/6\mp\ldots$, multiply through and calculate enough terms to get all the terms with negative powers.

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    Exactly what I would have said, except that I would have recommended finding the Laurent series expansion directly, by the method of long division of power series. It's exactly like long division of polynomials, except that you write the terms with the degrees *increasing* to the right.2013-03-28