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

[gammy]

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.

 

[Dick]

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.

 

[AKG]

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.

 

[gammy]

thank you very much for your help.

 

[gammy]

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.

 

[Dick]

Use Green's theorem on it.

 

[gammy]

Thank you sir! Solved! Not sure how to update thread title.