Contour Integral of |z| = 2 using Cauchy's Formula

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SUMMARY

The discussion focuses on evaluating the contour integral of the function \( \frac{1}{z^3} \) over the circle defined by \( |z| = 2 \) using Cauchy's Integral Formula. The integral simplifies to zero due to the higher order derivatives of the function being zero when applying the formula with \( n = 2 \), \( a = 0 \), and \( f(z) = 1 \). The parametrization of the contour can be achieved using \( z(t) \), but the conclusion remains that the integral evaluates to zero.

PREREQUISITES
  • Cauchy's Integral Formula
  • Complex analysis fundamentals
  • Parametrization of curves in the complex plane
  • Understanding of contour integrals
NEXT STEPS
  • Study the applications of Cauchy's Integral Formula in complex analysis
  • Learn about higher order derivatives in the context of complex functions
  • Explore parametrization techniques for different contours in the complex plane
  • Investigate the implications of contour integrals in evaluating residues
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Students of complex analysis, mathematicians focusing on contour integration, and anyone seeking to deepen their understanding of Cauchy's Integral Formula and its applications.

muzak
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Homework Statement


|z| = 2, \oint\frac{1}{z^3}


Homework Equations


Cauchy's Integral Formula
http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution


Seems like a simple application of the general formula on the wiki page with n = 2, a = 0, and f(z) = 1. The higher order derivatives just yield zero, making the integral zero. Just asking for verification for a friend.
 
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You're using a circle of radius 2 as a contour. You could parametrize it with some ##z(t)##, and then evaluate it. Think about ##e##, and see what you get.
 

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