- Let \(\scriptstyle \sum _ { n = - \infty } ^ { \infty } a _ { n } z ^ { n }\) be the Laurent series expansion of \(\scriptstyle { \frac { 1 } { \sin z } }\) on the annulus \(\left\{ z \in \mathbb { C } : \pi < | z | < 2 \pi \right\}\) . Evaluate the coefficients \(a _ { n }\) for \(n < 0\)
Solution
For \(m\ge1\), the negative coefficients are \[ \boxed{ a_{-1}=-1,\qquad a_{-2k}=0\ (k\ge1),\qquad a_{-(2k+1)}=-2\pi^{2k}\ (k\ge1). } \]
1. A negative Laurent coefficient is a residue sum inside any circle with \(\pi<r<2\pi\).
Fix \(r\) with \(\pi<r<2\pi\). The Laurent coefficient formula gives, for \(m\ge1\), \[ a_{-m} =\frac1{2\pi i}\int_{|z|=r} \frac{1/\sin z}{z^{-m+1}}\,dz =\frac1{2\pi i}\int_{|z|=r}\frac{z^{m-1}}{\sin z}\,dz. \] The zeros of sine are exactly the integer multiples of \(\pi\), and they are simple because \(\cos(k\pi)=(-1)^k\ne0\). Inside \(|z|<r\) the only such points are \(0,\pi,-\pi\). Therefore the residue theorem expresses \(a_{-m}\) as the sum of the residues of \(z^{m-1}/\sin z\) at those three points.
2. The residues at \(\pm\pi\) give the parity pattern.
At \(z=k\pi\), the residue of \(1/\sin z\) is \[ \frac1{\cos(k\pi)}=(-1)^k. \] Thus at both \(\pi\) and \(-\pi\) the residue is \(-1\), and hence \[ \operatorname{Res}_{\pi}\frac{z^{m-1}}{\sin z} +\operatorname{Res}_{-\pi}\frac{z^{m-1}}{\sin z} =-\pi^{m-1}-(-\pi)^{m-1}. \] If \(m\) is even, then \(m-1\) is odd and this sum is zero. If \(m\) is odd, it is \(-2\pi^{m-1}\).
3. The origin contributes only when \(m=1\).
Near zero, \[ \sin z=z+O(z^3), \qquad \frac{z^{m-1}}{\sin z}=z^{m-2}(1+O(z^2)). \] For \(m=1\) this has residue \(1\). For every \(m\ge2\) it is holomorphic at zero, so its residue there is zero.
Combining this with step 2 gives \[ a_{-1}=1-1-1=-1. \] For even \(m\ge2\), \(a_{-m}=0\), while for odd \(m\ge3\), \(a_{-m}=-2\pi^{m-1}\). Writing \(m=2k\) or \(m=2k+1\) gives exactly the displayed formulas.