What Are the Residues and Poles of a Function?

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SUMMARY

The discussion centers on evaluating the integral \(\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{w(\zeta)}\frac{w(\zeta)-w(z)}{\zeta -z}d\zeta\) where \(f\) is holomorphic across the complex plane and \(w(z)\) is a polynomial of degree \(n\) with distinct zeroes. The solution manual indicates that the integral equals \(\sum_{j=1}^n Res\left(\frac{f(\zeta)}{w(\zeta)}\frac{w(\zeta)-w(z)}{\zeta -z}, \zeta_j\right)\) with \(\zeta_j\) being the zeroes of \(w\). A key point raised is whether \(\zeta = z\) also constitutes a pole in the integrand, which would not contribute to the integral if excluded from the contour.

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quasar987
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Here I must evaluate

[tex]\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{w(\zeta)}\frac{w(\zeta)-w(z)}{\zeta -z}d\zeta[/tex]

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

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

where [itex]\zeta_j[/itex] are the n zeroes of w. But isn't [itex]\zeta = z[/itex] also a pole for the integrand?
 
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If the contour excludes [itex]\zeta = z[/itex] then the pole will not contribute to the integral.
 

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