Residue theorem with contour integrals: $ \int_{\gamma} \frac{1}{z^{2}\sin(z)} dz$ where $\gamma(t) = e^{it}$ and $ 0 \leq t \leq 2\pi$

Question

I want to evaluate the integral $$ \int_{\gamma} \frac{1}{z^{2}\sin(z)} dz$$

where $\gamma(t) = e^{it}$ and $ 0 \leq t \leq 2\pi$ using the Residue theorem.

I've tried expanding sin(z) with Taylor expansion but I don't seem to be getting anywhere.

Answer 1

Taylor series shows the pole is of order 3 at point $z=0$, Since $$z^2 \sin z=z^3 (1-z^2/6+\cdots)$$ So we should use this formula to obtain residue: $$ a_{-1}=\frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)]_{z=z_0}$$

for $m=3$ and $z_0=0$.

Answer 2

$$ \oint_{\gamma} \frac{1}{z^{2}\sin(z)} dz= \oint_{\gamma} \frac{1}{z^{3}(1-z^2/3!+z^4/5!-\cdots)} dz$$ $$ \phantom{\oint_{\gamma} \frac{1}{z^{2}\sin(z)} dz}= \oint_{\gamma} \frac{(1+z^2/3!+7z^4/360-\cdots)}{z^{3}} dz$$ $$ \phantom{\oint_{\gamma} \frac{1}{z^{2}\sin(z)} dz}= \oint_{\gamma} {\left(\frac{1}{z^3}+\frac{1}{6z}+\frac{7z}{360}-\cdots\right)} dz$$ $$ \phantom{\oint_{\gamma} \frac{1}{z^{2}\sin(z)} dz}= 2\pi i\cdot \frac{1}{6}$$ $$ \phantom{\oint_{\gamma} \frac{1}{z^{2}\sin(z)} dz}=\frac{\pi i}{3}.$$