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
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
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.