TITLE: Suppose $\sum_{k=-\infty}^{\infty}a_kz^k$ and $\sum_{-\infty}^{\infty}b_kz^k$ converge to $1/\sin(\pi z)$. Find $b_k-a_k$.
QUESTION [5 upvotes]: Suppose that the Laurent series $\sum_{k=-\infty}^{\infty}a_kz^k$ converges to $1/\sin(\pi z)$ when $0<|z|<1$, and suppose that the Laurent series $\sum_{k=-\infty}^{\infty}b_kz^k$ converges to $1/\sin(\pi z)$ when $1<|z|<2$. Find, for every integer $k$, a simple expression for the difference $(b_k-a_k)$.

I don't quite know where to start with this. I was thinking that $1/\sin(\pi z)$ has simple poles at $z=\pm 1$, so we might be able to say $(z^2-1)\sum_{k=-\infty}^{\infty}a_kz^k$ and $(z^2-1)\sum_{k=-\infty}^{\infty}b_kz^k$ both converge for $0<|z|<2$, but I'm not quite sure.
I was also considering using Euler's product representation for sine somehow, but I couldn't quite figure out how.
Any help is greatly appreicated (hints or tips preferred). Thanks

REPLY [3 votes]: Hint: Note that $$\displaystyle f(z) = \frac{\pi}{\sin \pi z} - \frac{1}{z} + \frac{1}{z-1}+\frac{1}{z+1}$$ is holomorphic in the disc $D'(0,2)$ and can be expanded in Taylor Series, with the formula for the $n-$th coefficient $f_n = \dfrac{1}{n!}\left.\dfrac{d^nf(z)}{dz^n}\right\vert_{z = 0}$. You know the Laurent series for the $\displaystyle -\frac{1}{z} + \frac{1}{z-1}+\frac{1}{z+1}$ part, that gives you Laurent series for $\dfrac{\pi}{\sin \pi z}$.