I have to evaluate the following contour integral, along the positively oriented contour $C:|z-3|=2$
$$\oint\frac{4i(z^2+4)}{z(z^2-16)}\sin{\Bigg(\frac{5\pi}{z^2+4}\Bigg)}dz$$
Attempt:
since $z(z^2-16)=0$ is true if and only if $z=0$ or $z=\pm4$ but we can ignore everything but $4$ since it's the only one that's inside the contour.
so $$\oint\frac{\frac{4i(z^2+4)}{z(z+4)}\cdot\sin{\Big(\frac{5\pi}{z^2+4}\Big)}dz{}}{(z-4)}$$
By Cauchy Integral formula.
$$f(\zeta)=\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{z-\zeta}dz$$
where $\gamma$ is a closed oriented curve.
$$f(z)=\frac{4i(z^2+4)}{z(z+4)}\sin{\Bigg(\frac{5\pi}{z^2+4}\Bigg)}dz$$ $\zeta=4$ so
$$f(4)=\frac{4i\cdot20}{4\cdot8}\sin{\Bigg(\frac{5\pi}{20}\Bigg)}dz\Rightarrow f(4)=\frac{5i}{2}\sin{}\frac{\pi}{4}$$ Giving us the final answer as
$$I=2\pi if(4)=-\frac{5}{\sqrt{2}}$$
I'm not sure if this is correct and I was also wondering if there's any other way to approach this?