 Solution

Laurent's theorem states that a holomorphic function on $R<|z-a|<S$ has a unique expansion ${\sum }_{n\in \mathbb{Z}}{a}_n(z-a{)}^n$, converging absolutely there and uniformly on compact subannuli, withHere $\sin z=z(1-z^2/6+\cdots )$ has a simple zero at zero and no other zero in $|z|<\pi$. Hence $1/\sin z$ has only a simple pole there, giving the stated series with no powers below $z^{-1}$. Multiplying by $\sin z$ and comparing the constant, linear and quadratic coefficients yields

Use the coefficient integrals on the two circles. The annular region between them contains exactly the simple poles $\pi ,-\pi$, and the residue of $1/\sin z$ at each is $1/\cos (\pm \pi )=-1$. Apply the residue theorem to $1/(z^{n+1}\sin z)$ between the circles:For lower indices, set $c_n=0$ because the inner Laurent series has no such terms; the same contour argument then still applies. Since $f$ is odd, its even-index coefficients vanish. A complete explicit list for the requested range isThe subtraction accounts for the two poles crossed; using the inner coefficients unchanged outside their annulus would miss precisely these terms.