Integration along the contour

Question

I totally don't get this solution. What is the integral formula this solution refers to? And can someone provide a more detailed answer to this problem? Many thanks!

Answer 1

I believe this solution refers to Cauchy's Integral Formula, which states that

$$f^n(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}} dz$$

For infinitely differentiable $f$.

In this case, $n = 1$. Thus, by rearranging, we get

$$\oint_\gamma \frac{f(z)}{(z-a)^{2}} = 2\pi i f'(a)$$

In this case, your $a = \pi$ and $f(z) = e^{iz} \sin z$.