Complex Integral

The function $f(z) = \dfrac{\cos{z}}{(z-i)^3}$ has a pole of order $n=3$ at $z=i$, which is inside the countour of integration. This is because $(z-i)^3f(z) = \cos{z}$ is holomorphic (i.e., differentiable) and non-zero around $z=i$.

By the residue theorem, this integral equals $2 \pi \text{Res}(f,i)$, where $\text{Res}(f,i)$ is the residue of $f$ at $z=i$.
 
To calculate this residue, we can apply the formula:
$$\text{Res}(f,i) = \dfrac{1}{(n-1)!} \lim_{z\rightarrow i} \dfrac{d^{n-1}}{dz^{n-1}}(z-i)^nf(z)$$ where $n=3$.

This gives
$$\text{Res}(f,i) = \dfrac{1}{2} \lim_{z\rightarrow i} \dfrac{d^{2}}{dz^{2}}\cos{z} = -\dfrac{1}{2}\cos{i} = -\dfrac{1}{2}\left( \dfrac{e^{-1}+e}{2}\right) = - \dfrac{e^2+1}{4e}$$