Evaluating Complex Integrals: From Pi to i & Around Unit Circle & Square

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SUMMARY

The discussion focuses on evaluating the complex integral of e^(iz) from π to i, utilizing various paths including straight lines and segments along the coordinate axes. The participant expresses confusion regarding the conversion of z to x + iy and the implications for bounds. Additionally, they explore the integral around the unit circle, concluding that it equals 2πi, and seek clarification on evaluating the integral around a square path with vertices at ±1 on both axes, ultimately arriving at the answer of 8i through parametrization.

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of contour integration
  • Familiarity with parametrization of paths in the complex plane
  • Knowledge of the Cauchy Integral Theorem
NEXT STEPS
  • Study complex contour integration techniques
  • Learn about the Cauchy Integral Formula
  • Explore parametrization methods for complex paths
  • Investigate the properties of analytic functions in complex analysis
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Students of complex analysis, mathematicians working on contour integration, and anyone interested in evaluating complex integrals in various paths.

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Homework Statement



integral(e^(iz)dz) with bounds from pi to i.
I need to evaluate using : along the straight line joining the limits, along segments of the coordinate axes joining the limits, and estimate using integral <= L*M and compare the estimates with the values.

I am totally confused on how to do this. Do I convert z to x+iy and do the bounds change?



I have another integral with z conjugate about the unit circle. I think I have an idea on this one on converting to x-iy and seeing that this will equal 2ipi.



I am also asked to evaluate this same integral taken in the positive sense about the square with x=+-1, y=+-1. I know this involves 4 segments, but my problem is finding this 4 segments. I know the answer is 8i, but how to get there?
 
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You can parametrize the path from pi to i, e.g by taking:

z(t) = pi + (i-pi)t

The starting point is then at t = 0 and the end point at t = 1.

integral f(z)dz along the path then means:

f(z(t))d(z(t)) from t = 0 to t = 1.

We have dz = (i-pi) dt

and

f(z(t)) = exp[i pi - (i pi + 1)t] = -exp[-(i pi + 1)t]
 
That makes a lot of sense actually.
 

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