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V0ODO0CH1LD

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## Homework Statement

Calculate the complex integral along the closed path indicated:

$$ \oint_C\frac{\sin{z}}{z^2+\pi^2}dz,\,\,|z-2i|=2.$$

## Homework Equations

$$ \sin{z}=\frac{e^{iz}-e^{-iz}}{2i} $$

$$ e^{iz}=e^{i(x+iy)}=e^{-y+ix}=e^{-y}(\cos{x}+i\sin{x}) $$

## The Attempt at a Solution

I really don't know what to do here.. Everything I tried led me to a dead end. Is there a clever substitution to be made? I tried substituting ##z=x+iy##, I tried ##z=e^{it}+2i## and even tried expanding ##\sin{z}##, but it got me nowhere. Any help is appreciated. Thanks!