Problem of the Week #301 - June 28, 2022

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SUMMARY

The discussion centers on the mathematical proof that if a function \( f \) is holomorphic in an open subset \( \Omega \subset \mathbb{C} \), then for all closed contours \( \Gamma \) within \( \Omega \), the integral \( \oint_{\Gamma} \overline{f(z)}f’(z)\, dz \) is purely imaginary. This conclusion is derived from properties of holomorphic functions and their derivatives. The proof leverages the Cauchy-Riemann equations and the nature of complex conjugates in integration.

PREREQUISITES
  • Understanding of holomorphic functions in complex analysis
  • Familiarity with closed contours in the complex plane
  • Knowledge of the Cauchy-Riemann equations
  • Basic principles of complex integration
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  • Study the properties of holomorphic functions in detail
  • Explore the Cauchy Integral Theorem and its applications
  • Learn about complex conjugates and their role in integrals
  • Investigate the implications of purely imaginary integrals in complex analysis
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Mathematicians, students of complex analysis, and anyone interested in advanced calculus and integrals in the complex plane.

Euge
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Here is this week's problem!

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Prove that if $f$ is holomorphic in an open subset $\Omega\subset \mathbb{C}$, then for all closed countours $\Gamma$ in $\Omega$, the integral $\oint_{\Gamma} \overline{f(z)}f’(z)\, dz$ is purely imaginary.
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Here is my solution.

The differential form ##\overline{f(z)}f'(z)\, dz = \overline{f(z)}\,df(z)##, so ##2\operatorname{Re}(\overline{f(z)}f'(z)\, dz) = \overline{f}\, df + f\, d\overline{f} = d(f\overline{f})##, an exact differential. Therefore $$2\operatorname{Re} \oint_\Gamma \overline{f(z)}f'(z)\, dz = \oint_\Gamma 2\operatorname{Re}(\overline{f(z)}f'(z)\, dz) = 0$$ and the result follows.
 
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