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[parent] zero as contour integral (Corollary)

Suppose that $ f$ is a complex function which is defined in some open set $ D \subseteq \mathbb{C}$ which has a simple zero at some point $ p \in D$. Then we have

$\displaystyle p = {1 \over 2 \pi i} \oint_C {z f'(z) \over f(z)} \, dz $
where $ C$ is a closed path in $ D$ which encloses $ p$ but does not enclose or pass through any other zeros of $ f$.

This follows from the Cauchy residue theorem. We have that the poles of $ f'/f$ occur at the zeros of $ f$ and that the residue of a pole of $ f'/f$ is $ 1$ at a simple zero of $ f$. Hence, the residue of $ z f'(z) / f(z)$ at $ p$ is $ p$, so the above formula follows from the residue theorem.



"zero as contour integral" is owned by rspuzio. [ full author list (2) ]
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Cross-references: residue, poles, Cauchy residue theorem, pass through, closed path, point, simple zero, open set, complex function

This is version 2 of zero as contour integral, born on 2007-03-03, modified 2007-03-03.
Object id is 9008, canonical name is ZeroAsContourIntegral.
Accessed 504 times total.

Classification:
AMS MSC: 30E20 (Functions of a complex variable :: Miscellaneous topics of analysis in the complex domain :: Integration, integrals of Cauchy type, integral representations of analytic functions)
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