Calculating residue of a line integral

Question

My book has a confusing residue calculation:

$$\int_c \frac{4-3z}{z^2-z}dz$$

$$\int_c \frac{4-3z}{z(z-1)}dz$$

So when evaluating the residue at z = 0 $$Res_{z=0} [ \frac{4-3z}{z(z-1)}]$$

I don't understand how subbing z=0 gives us this? Shouldn't the denominator just be 0 since when z=0 the whole denom = 0?

$$= [\frac{4-3z}{z-1}]_{z=0} = -4$$

Answer 1

The notation $\operatorname{res}_{z=a}f(z)$ means “residue of $f$ at $a$”. So, $\operatorname{res}_{z=0}\frac{4-3z}{z(z-1)}$ stands for the residue of $\frac{4-3z}{z(z-1)}$ at $0$ (which is indeed $-4$).