I have to find: $$I=\oint_{\gamma}\frac{dz}{(z^2+4)^2}.$$
$\gamma$ in this case is a circular curve defined by $|z-i|=2$, which is a circle centered at $i$ with radius $2$. It is clear that the singularity here is $+2i$ and $-2i$, the latter of which is outside the circle. When a singularity is outside a circle its $0$ so do I just evaluate the one within the circle? By that would I just add them?
The residue theorem gives that the integral of a meromorphic function $f(z)$ on a simple closed contour $\gamma$, provided that $f(z)$ is regular in a small neighbourhood of $\gamma$, equals $2\pi i$ times the sum of the residues of $f(z)$ at the singularities inside $\gamma$. So we have:
$$I=\oint_{|z-i|=2}\frac{dz}{(z^2+4)^2} = 2\pi i\,\operatorname{Res}\left(\frac{1}{(z^2+4)^2},z=2i\right) $$ and since $z=2i$ is a double pole for the given function it follows that: $$ I = 2\pi i\cdot\lim_{z\to 2i}\frac{d}{dz}\frac{(z-2i)^2}{(z^2+4)^2}=2\pi i\cdot\lim_{z\to 2i}\frac{d}{dz}\frac{1}{(z+2i)^2}=-4\pi i\cdot\lim_{z\to 2i}\frac{1}{(z+2i)^3}=\color{red}{\frac{\pi}{16}}.$$