Understanding the Cauchy Integral Formula and Evaluating Complex Integrals

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SUMMARY

The discussion focuses on evaluating the complex integral using the Cauchy Integral Formula (CIF) for the function \(\frac{e^{2iz}}{(3z-1)^2}\) with a singularity at \(z = \frac{1}{3}\) inside the contour \(|z| = 4\). The initial solution provided was \(2\pi i e^{2/3i}\), but the correct answer is \(-\frac{4\pi}{9} e^{2/3i}\). The discrepancy arises from the need to apply the generalized Cauchy Integral Formula due to the repeated root in the denominator, which introduces a factor of \(\frac{1}{n!}\) where \(n\) is the order of the pole.

PREREQUISITES
  • Cauchy Integral Formula (CIF)
  • Complex analysis fundamentals
  • Understanding of singularities in complex functions
  • Knowledge of evaluating residues and poles
NEXT STEPS
  • Study the generalized Cauchy Integral Formula for higher-order poles
  • Learn about residue calculus in complex analysis
  • Explore examples of complex integrals involving repeated roots
  • Review the properties of exponential functions in complex variables
USEFUL FOR

Students of complex analysis, mathematicians working with integrals, and anyone seeking to deepen their understanding of the Cauchy Integral Formula and its applications in evaluating complex integrals.

NewtonianAlch
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Homework Statement


Evaluating using CIF.

|z| = 4

Integral {\frac {{{\rm e}^{2\,iz}}{\it dz}}{ \left( 3\,z-1 \right) ^{2}}}



The Attempt at a Solution



So the singularity here is z = 1/3 which is inside the circle.

Therefore using the formula 2\,i\pi \,f and substituting in the z = 1/3

We get f(1/3) = exp(2/3i)

So I get the answer of 2*Pi*i(exp(2/3i))

However the answer given is (-4Pi/9)(exp(2/3i))

I'm thinking it has something to do with the repeated root, but I'm not sure.
 
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Found out that when dealing with powers, a generalised CIF has to be used. Now however I get -4Pi(exp(2/3i))

Where did that 1/9 come from?
 

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