Proving the Area Enclosed by a Simple Closed Path using Complex Analysis

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Homework Help Overview

The discussion revolves around proving that the area enclosed by a positively oriented simple closed piecewise smooth path \( C \) can be expressed using complex analysis, specifically through the integral \( (1/2i) \int_{C} \bar{z} dz \). Participants are exploring the relationship between complex integrals and area calculation, referencing Green's theorem as a potential method for this proof.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the parametrization of the curve \( C \) and the use of functions \( f(t) \) and \( g(t) \) to facilitate the proof. There is mention of breaking the contour into real and imaginary parts and applying Green's theorem to relate the area to integrals involving \( x \) and \( y \). Some participants express uncertainty about how to proceed with the proof and seek guidance on evaluating specific integrals.

Discussion Status

The conversation includes various approaches to the problem, with some participants suggesting the use of Green's theorem and others working through the implications of their calculations. While one participant indicates they have reached a solution, there remains a focus on understanding the steps leading to that conclusion, with no explicit consensus on the final proof method.

Contextual Notes

Participants are navigating the complexities of contour integration and the application of Green's theorem, with some expressing uncertainty about the assumptions and definitions involved in their reasoning. There is an acknowledgment of the piecewise smooth nature of the curve \( C \) as a relevant constraint.

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Homework Statement


C = positively oriented simple closed piecewise smooth path

Prove that:

(1/2i)*\int_{C}\bar{z}dz

is the area enclosed by C.

Homework Equations



*I know that the curve C is piecewise smooth so that it can be broken up into finitely many pieces so that each piece is smooth.

*Cauchy's integral formula

The Attempt at a Solution



I think that I want to let some function f(t) be a continuous complex-valued function on the path C. Let g(t) be a parametrization of the curve. Frankly, I'm not sure where to go from here. I've just started doing contour integration problems and I have problems knowing where to start with proofs.

I'm just hoping someone can point me in the right direction on where I might want to begin. Thanks.
 
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You can compute the area inside a contour using Green's theorem as (1/2) integral xdy-ydx. Break your complex contour into real and imaginary parts.
 
A consequence of Green's Theorem is that the area enclosed by C is:

(1/2)\int_C xdy - ydx

which is

(1/2)\int _0 ^1 x(t)y'(t)dt - y(t)x'(t)dt

where g : [0,1] -> C, g(t) = x(t) + iy(t) is your parametrization of C.
 
thank you very much for your help.
 
Thank you again for the assistance. I have almost gotten to the conclusion that I need but I'm not sure how to proceed.

I have shown that:

(1/2)\int_{0}^{1} \bar{z}dz = (1/2)\int_{0}^{1} x(t)y'(t)dt-y(t)x'(t)dt + (1/2i)\int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt

I am assuming that the integral \int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt is equal to zero because if so I will obtain the result I am looking for, namely that:

(1/2)\int_{0}^{1} \bar{z}dz = (1/2)\int_{0}^{1} x(t)y'(t)dt-y(t)x'(t)dt

I'm not sure how to evaluate \int_{0}^{1} x(t)x'(t)dt-y(t)y'(t)dt.

Any ideas would be much appreciated.
 
Use Green's theorem on it.
 
Thank you sir! Solved! Not sure how to update thread title.
 

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