Evaluate the integral by means of the Cauchy residue theorem.
$$\oint_{|z|=1} \dfrac{1}{z^{2} \sin z} dz$$.
So Cauchy's Residue Theorem states:
If $\Gamma$ is a simple closed positively oriented contour and $f$ is analytic inside and on $\Gamma$ except at the points $z_{1},z_{2},…,z_{n}$ inside $\Gamma$, then
$$\int_{\Gamma} f(z)dz = 2\pi i \sum_{j=1}^{n} Res (z_{j})$$
where Res denotes the residue of $f$ at $z_{0}$. I thought that I could go about this problem by letting $f= \dfrac{1/ \sin z}{(z-0)^{2}}$ and then taking the derivative of $\dfrac{1}{ \sin z}$ at $z_0=0$ to get the residue of $f$ at $0$, but then I get that the residue of $f$ at $0$ is $1$, and then using Cauchy's Residue Theorem, I get that
$$\oint_{|z|=1} \dfrac{1}{z^{2} \sin z} dz = 2\pi i$$
which is different from the answer in the book. I'd like to know where I'm going wrong.
Best Answer
You need the residue of $f(z)$ at $z=0$. Expanding as a Laurent series gives $$f(z)=z^{-2}\left(z-\frac{z^3}6+\cdots\right)^{-1} =z^{-3}\left(1-\frac{z^2}6+\cdots\right)^{-1} =z^{-3}\left(1+\frac{z^2}6+\cdots\right) =z^{-3}+\frac{1}{6z}+\cdots$$ giving the residue as $1/6$. Therefore the integral is $\pi i/3$.