$$ \int _ {|z|=2} \frac { dz} {(z-4)(z^3-1)} $$
What I've done now is the following.
$f$ has isolated singularities at $z=4$, $1$, $\exp(\pi i/3)$, $\exp(-\pi i / 3)$
$$ \int _ {|z|=2} \frac { dz} {(z-4)(z^3-1)} = 2 \pi i (\operatorname{Res}(f;1) + \operatorname{Res}(f;\exp(\pi i / 3) + \operatorname{Res}(f;\exp(-\pi i / 3) ) $$
Does it make sense?
And I have lost my way..
In my calculation, Res(f;1) = $\frac 1 {-9}$. But I don't calculate the following 2 values.
The poles are at $z=e^{\pm i 2 \pi/3}$; compute the residues from there.
$$\text{Res}_{z=e^{i 2\pi/3}} \frac{1}{(z-4)(z^3-1)} = \frac{1}{(e^{i 2 \pi/3}-4) 3 e^{i 4 \pi/3}}$$
$$\text{Res}_{z=e^{-i 2\pi/3}} \frac{1}{(z-4)(z^3-1)} = \frac{1}{(e^{-i 2 \pi/3}-4) 3 e^{-i 4 \pi/3}}$$