Evaluate $\int_{0}^{2\pi} \sin(\frac{\pi}{6} - 2\text{exp}(i\theta)) d\theta$

Question

In one of my exercise sheets, I am asked to find $\int_{0}^{2\pi}\sin(\frac{\pi}{6}-2\exp(i\theta))d\theta$

This follows a question asking to derive a form of Cauchy's theorem:

$f(a)=\frac{1}{2n\pi}\int_{0}^{2n\pi}f(a+r\exp{i\theta})d\theta$

But I can't even figure out how to start.

All help appreciated, please help me understand this!

Thank you.

EDIT:

So I understood part 1, deriving $f(a)=\frac{1}{2n\pi}\int_{0}^{2n\pi}f(a+r\exp{i\theta})d\theta$

Now I need to evaluate $\int_{0}^{2\pi}\sin(\frac{\pi}{6}-2\exp(i\theta))d\theta$

Answer 1

Let $a+r\exp{i\theta}=\frac{\pi}{6}+2\exp{i\theta}$, so $a=\frac{\pi}{6}$

Now we have $$2\pi f\left(\frac{\pi}{6}\right)=\int_{0}^{2\pi}\sin\left(\frac{\pi}{6}+r\exp{i\theta}\right)d\theta$$

So $$\int_{0}^{2\pi}\sin\left(\frac{\pi}{6}+r\exp{i\theta}\right)d\theta=2\pi\frac{1}{2}=\pi$$

as $f(a)=\sin(a)$