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Prove that if a differentiable curve $g:[0,1] \to \mathbb{C}$ (complex plane) parametrizes counterclockwise the boundary of an open set $O$ in $\mathbb{C}$, then under suitable conditions area of $O$ is $$ {1 \over 2i } \int_g \overline{z} dz $$ computed over the boundary.

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1 Answer 1

Write your contour integral out explicitly using the parameterization $g(t)$ and then use Green's theorem.

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z = g(t) = x(t) + i y(t) then what I have to prove becomes 1/2i* integral [xx' +yy' + i (xy' -yx')] not sure what to do from here –  user6161 Jan 23 '11 at 20:50
    
use Green's theorem on the real and imaginary parts, and then apply the Cauchy-Riemann equations to the resulting two terms. One will end out being zero, the other will be what you want. –  Zarrax Jan 23 '11 at 21:03
    
so if i m to continue from my previous steps, i get 1/2i*int(xx'+yy') + 1/2*int(xy'-yx'). the second integral i think is the area of O by green's theorem. is that right? and if so, how do i prove the first integral becomes 0. is the first integral equal to 1/2i * double int(dy/dx - dx/dy) dA? i feel like i m wrong all over the place. thanks so much for your help! –  user6161 Jan 23 '11 at 22:12
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for example $\int(xx' + yy')$ is $\int(Pdx + Qdy)$ where $P$ is $x$ and $Q$ is $y$. ${\partial Q \over \partial x}$ and ${\partial P \over \partial y}$ are thus both zero and using Green's Theorem the integral of the first term is zero. Now try a similar thing in the second term and see what happens. Btw you don't even need the Cauchy-Riemann equations, I don't know why I said that. –  Zarrax Jan 23 '11 at 22:42
    
oh! u r a savior! i can be so silly sometimes :-) thanks a lot! –  user6161 Jan 23 '11 at 23:10

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