Let $C$ be the circle $\mid \:z \mid = 6$ traced one lap counterclockwise. Evaluate: $$\int_{C} \frac{\cos(z)}{(z+i)^3}\;dz$$
Since $\cos(z)$ is analytic, my solution was to use cauchys integral formula; $$f^n(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz$$ where $f(z) = \cos(z)$ and $z_0 = -i$.
$ n=2$ in this case, so we get $$f''(z_0) = \frac{1}{\pi i} \int_{C} \frac{f(z)}{(z-z_0)^3}dz \; \iff \int_{C} \frac{f(z)}{(z-z_0)^3}dz = f''(z_0) \pi i$$
$f''(z_0)= -\cos(-i) = -\cos(i)$
$\therefore \int_{C} \frac{\cos(z)}{(z+i)^3}dz =-\cos(i) \pi i$
Just wanted to make sure this is correct.