# Complex trigonometric integral

1. Nov 21, 2014

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data
Calculate the complex integral along the closed path indicated:
$$\oint_C\frac{\sin{z}}{z^2+\pi^2}dz,\,\,|z-2i|=2.$$
2. Relevant equations
$$\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$$
$$e^{iz}=e^{i(x+iy)}=e^{-y+ix}=e^{-y}(\cos{x}+i\sin{x})$$
3. 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!

2. Nov 21, 2014

### ShayanJ

You should use calculus of residues!

3. Nov 21, 2014

### Zondrina

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

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

4. Nov 21, 2014

### V0ODO0CH1LD

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.

5. Nov 21, 2014

### ShayanJ

Check here!

6. Nov 21, 2014

### Zondrina

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