Interesting complex variables problem

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SUMMARY

The discussion centers on proving the area of a bounded domain \(\Omega\) in the complex plane, represented by the boundary curve \(z = z(t)\) for \(a \leq t \leq b\). The area \(A(\Omega)\) is established as \(A(\Omega) = \frac{1}{2}\int^b_a |z(t)|^2 \text{Im}\left(\frac{z'(t)}{z(t)}\right)dt\). Participants highlight the connection to Green's theorem, noting that the area can also be expressed as \(A(\Omega) = \frac{1}{2}\int_C x dy - y dx\) when \(z(t) = x(t) + iy(t)\). This demonstrates the equivalence of the two area calculations.

PREREQUISITES
  • Understanding of complex variables and functions
  • Familiarity with Green's theorem in vector calculus
  • Knowledge of the properties of integrals and imaginary components
  • Basic proficiency in parameterization of curves in the complex plane
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  • Study the application of Green's theorem in calculating areas
  • Explore complex analysis concepts, particularly contour integration
  • Learn about the properties of complex derivatives and their geometric interpretations
  • Investigate the relationship between complex functions and real-valued integrals
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Homework Statement


Let \Omega be a bounded domain in C whose boundary is a curve z = z(t), a<=t<=b, and let A(\Omega) be the area of \Omega. Prove that

A(\Omega) = \frac{1}{2}\int^b_a |z(t)|^2 Im(\frac{z&#039;(t)}{z(t)})dt


Homework Equations





The Attempt at a Solution


Not even sure where to start on this. Any tips?
 
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Yes. A(\Omega) = \frac{1}{2}\int_C x dy-y dx. That's a well known expression for calculating area using Green's theorem. If you express z(t)=x(t)+iy(t), that's what your expression reduces to.
 
Ahh right, thank you
 

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