Complex trigonometric integral

[V0ODO0CH1LD]

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!

 

[ShayanJ]

You should use calculus of residues!

 

[STEMucator]

The denominator $z^2 + \pi^2$ has singularities at $z = \pm i \pi$.

Do these lie within the positively oriented contour $|z - 2i| = 2$?

 

[V0ODO0CH1LD]

The denominator $z^2 + \pi^2$ has singularities at $z = \pm i \pi$.

Do these lie within the positively oriented contour $|z - 2i| = 2$?

Yes, $i\pi$ lies within the contour, which means that the integral for any closed path around $i\pi$ would wield the right answer, but that didn't help me much. I still don't know how to calculate the integral.

 

[ShayanJ]

Check here!

 

[STEMucator]

$$\oint_C f(z) \space dz = (2 \pi i) \times \space \text{Res}[f(z), i \pi]$$