What Are the Residues and Poles of a Function?

[quasar987]

Here I must evaluate

$$\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{w(\zeta)}\frac{w(\zeta)-w(z)}{\zeta -z}d\zeta$$

where f is holomorphic on the hole complex plane, where w(z) is a polynomial of degree n with all of its zeroes distinct (i.e. all n have multiplicity 1), and where C is a closed curve containing all the zeroes of w in its interior.

The solution manual says that the integral equals

$$\sum_{j=1}^n Res(\frac{f(\zeta)}{w(\zeta)}\frac{w(\zeta)-w(z)}{\zeta -z}, \zeta_j)$$

where $\zeta_j$ are the n zeroes of w. But isn't $\zeta = z$ also a pole for the integrand?

 

[Tide]

If the contour excludes $\zeta = z$ then the pole will not contribute to the integral.

 

[tacman]

Yes, \zeta = z is also a pole for the integrand. However, when evaluating the residue at \zeta = z, the term \frac{w(\zeta)-w(z)}{\zeta -z} will evaluate to 1 instead of 0, so it does not affect the overall sum. Therefore, the solution manual only includes the residues at the n zeroes of w, as those are the only terms that contribute to the sum.