$\int_c \frac{8-z}{z(4-z)} dz$

Question

I want to calculate the following contour integral: $$\int_c \frac{8-z}{z(4-z)} dz$$ where $C$ is the circle of radius $7$, center $0$, negatively oriented.

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Do I have to do this the long and tedious way or is there a nicer short cut?

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This has a pole at $0$ and $4$, so I need to calculate the residues for these two and take the line integral across the paramterisation of a circle moving clockwise of radius $7$, centered at the origin, that's all there is to it right?

Answer 1

Use the residue theorem to have:

$$ I = 2 \pi \mathrm{i} \left( \mathrm{Res}(f)_{z=0} + \mathrm{Res}(f)_{z=4} \right) = 2 \pi \mathrm{i} \left( 2 -1\right) = 2\pi \mathrm{i} $$

Cheers!