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Math Help - Complex- area enclosed by a polygon

  1. #1
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    Complex- area enclosed by a polygon

    Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,znis1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)Sh ow that the area enclosed =1/2Isummation[zkcongugate(zsub(k+1)-zk)].Interpret this sum as part of the approximating sum in the definition ofthe line integral about C of z congugatedz. I don't even know where to start. I understand where the 1/2I(z.....) comesfrom, but once the summation comes in, I'm lost.
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,znis1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)Sh ow that the area enclosed =1/2Isummation[zkcongugate(zsub(k+1)-zk)].Interpret this sum as part of the approximating sum in the definition ofthe line integral about C of z congugatedz. I don't even know where to start. I understand where the 1/2I(z.....) comesfrom, but once the summation comes in, I'm lost.

    Let \Gamma be polygon with its ordered vertices (the vertices you take in order as you travel around it) be z_1,z_2,...,z_n.
    Also say they are taken in a counterclockwise fashion.

    Let \Gamma_j = [z_j,z_{j+1}] be the line segment from z_j to z_{j+1} for j=1,2...,n-1. But for j=n define \Gamma_n = [z_n,z_1].

    Define f: \mathbb{C}\to \mathbb{C} by f(x+iy) = (-y+ix).
    Let R denote the interior of the polygon, then by Green's theorem:
    \frac{1}{2}\oint_{\Gamma} f = \frac{1}{2}\iint_R \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y}  = \iint_R 1 = \text{area}(R).

    Thus, the area is equal to: \frac{1}{2}\oint_{\Gamma}f = \frac{1}{2}\sum_{j=1}^n \int_{\Gamma_j}f = \frac{1}{2}\sum_{j=1}^{n-1} \left( \int_{[z_j,z_{j+1}]} f \right) + \frac{1}{2}\int_{[z_n,z_1]} f

    For j=1,2,...,n-1 define \gamma_j(t) = z_j + (z_{j+1}-z_j) t for 0\leq t\leq 1. Notice that \gamma_j(t) parametrizes [z_j,z_{j+1}].
    Thus, \int_{[z_j,z_{j+1}]} f = \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt.

    Now, f(x+iy) = -y + ix = i(x-iy), so f(z) = i\bar z, while \dot{\gamma}_j(t) = z_{j+1}-z_j.

    Therefore, \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt = i(z_{j+1}-z_j)\int_0^1 \overline{\gamma}_j(t)  dt = i(z_{j+1}-z_j)\bar z_j + \tfrac{1}{2}i|z_{j+1}-z_j|^2

    So if you add these up including [z_n,z_1] you get:
    \tfrac{1}{2}i( (z_2 - z_1)\bar z_1 + ... + (z_1 - z_n)\bar z_n) + \tfrac{1}{4}i(|z_1-z_2|+...+|z_n-z_1|)

    Work on simplying that expression now.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let \Gamma be polygon with its ordered vertices (the vertices you take in order as you travel around it) be z_1,z_2,...,z_n.
    Also say they are taken in a counterclockwise fashion.

    Let \Gamma_j = [z_j,z_{j+1}] be the line segment from z_j to z_{j+1} for j=1,2...,n-1. But for j=n define \Gamma_n = [z_n,z_1].

    Define f: \mathbb{C}\to \mathbb{C} by f(x+iy) = (-y+ix).
    Let R denote the interior of the polygon, then by Green's theorem:
    \frac{1}{2}\oint_{\Gamma} f = \frac{1}{2}\iint_R \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} = \iint_R 1 = \text{area}(R).

    Thus, the area is equal to: \frac{1}{2}\oint_{\Gamma}f = \frac{1}{2}\sum_{j=1}^n \int_{\Gamma_j}f = \frac{1}{2}\sum_{j=1}^{n-1} \left( \int_{[z_j,z_{j+1}]} f \right) + \frac{1}{2}\int_{[z_n,z_1]} f

    For j=1,2,...,n-1 define \gamma_j(t) = z_j + (z_{j+1}-z_j) t for 0\leq t\leq 1. Notice that \gamma_j(t) parametrizes [z_j,z_{j+1}].
    Thus, \int_{[z_j,z_{j+1}]} f = \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt.

    Now, f(x+iy) = -y + ix = i(x-iy), so f(z) = i\bar z, while \dot{\gamma}_j(t) = z_{j+1}-z_j.

    Therefore, \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt = i(z_{j+1}-z_j)\int_0^1 \overline{\gamma}_j(t) dt = i(z_{j+1}-z_j)\bar z_j + \tfrac{1}{2}i|z_{j+1}-z_j|^2

    So if you add these up including [z_n,z_1] you get:
    \tfrac{1}{2}i( (z_2 - z_1)\bar z_1 + ... + (z_1 - z_n)\bar z_n) + \tfrac{1}{4}i(|z_1-z_2|+...+|z_n-z_1|)

    Work on simplying that expression now.
    Ok, I follow this pretty well until the modulus comes in. Why exactly do we have a modulus?
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    Quote Originally Posted by kathrynmath View Post
    Ok, I follow this pretty well until the modulus comes in. Why exactly do we have a modulus?
    For z\in \mathbb{C} we have that z\bar z = |z|^2.
    The modulus comes in because we multiply complex numbers by their conjugates.
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  5. #5
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    Ok, I don't understand this as well as I thought. Why exactly are we using a segment from zj to zsub(j+1)? Is it because we can only evaluate the polygon on each side?

    Next, I don't understand the f(x+iy)=-y+ix

    The next part I understan up to a certain point. I see how Green's Theorem works, but I don't see where the 1/2 comes from.
    The next line following makes me all lost with the summations. I'm thinking it comes somewhat from the definition of a line integral, but I'm a bit confused.
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  6. #6
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    Quote Originally Posted by kathrynmath View Post
    Ok, I don't understand this as well as I thought. Why exactly are we using a segment from zj to zsub(j+1)? Is it because we can only evaluate the polygon on each side?
    Yes, \Gamma is the entire polygon. We have shown that \frac{1}{2}\oint_{\Gamma} f is the area of the polygon by Green's theorem. Now to evaluate \oint_{\Gamma} f we break up the polygon into each side and evaluate every side along those lines [z_j,z_{j+1}].

    Next, I don't understand the f(x+iy)=-y+ix
    I defined f: \mathbb{C}\to \mathbb{C} to be f(x+iy) = -y+ix. The reason why I defined it that way is because \tfrac{\partial f}{\partial x} = 1 \text{ and }\tfrac{\partial f}{\partial y} = -1 so \tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y} = 2. This differencial term is what appeas in Green's theorem.

    The next part I understan up to a certain point. I see how Green's Theorem works, but I don't see where the 1/2 comes from.
    Because we get (without the 1/2): \iint_R 2 = 2\cdot \text{area}(R).
    Therefore, \oint_{\Gamma} f is twice the area, we need to divide by 2 to get the actual area.
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  7. #7
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    ok I think I get all that, but is there a reason we choose a certain f or would this f(x+iy) be the case every time?
    Also, you say 1/2 of the integral f is the area of the polygon by Green's Theorem. Does this come from somewhere?
    Last edited by kathrynmath; May 1st 2009 at 03:38 PM.
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  8. #8
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    Ok now after I use Green's Theorem I'm trying to understand the summation stuff. I'm not following.
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  9. #9
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    Quote Originally Posted by kathrynmath View Post
    ok I think I get all that, but is there a reason we choose a certain f or would this f(x+iy) be the case every time?
    I choose f(x+iy) = -y+ix because it works. There are other functions that can give \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} = 1. This is a typical function to use for finding area by Green's theorem.

    Also, you say 1/2 of the integral f is the area of the polygon by Green's Theorem. Does this come from somewhere?
    Remember that \iint_R 1 = \text{area}(R).
    Therefore, \iint_R 2 = 2\text{area}(R).

    Quote Originally Posted by kathrynmath View Post
    Ok now after I use Green's Theorem I'm trying to understand the summation stuff. I'm not following.
    What part do you not understand? Which part?
    The first thing I did was break up \Gamma into line segments.
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    Quote Originally Posted by ThePerfectHacker View Post
    I choose f(x+iy) = -y+ix because it works. There are other functions that can give \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} = 1. This is a typical function to use for finding area by Green's theorem.


    Remember that \iint_R 1 = \text{area}(R).
    Therefore, \iint_R 2 = 2\text{area}(R).


    What part do you not understand? Which part?
    The first thing I did was break up \Gamma into line segments.
    Did you break up into line segments of length say z1 to z2, z2 to z3 and so on?
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    Quote Originally Posted by kathrynmath View Post
    Did you break up into line segments of length say z1 to z2, z2 to z3 and so on?
    When I write [z1,z2] I mean the line segment from z1 to z2. Now \Gamma consists of the line segments [z1,z2],[z2,z3],...,[zn,z1]. And that is all I did, I just integrate that function along each line segment.
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  12. #12
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    Quote Originally Posted by ThePerfectHacker View Post
    When I write [z1,z2] I mean the line segment from z1 to z2. Now \Gamma consists of the line segments [z1,z2],[z2,z3],...,[zn,z1]. And that is all I did, I just integrate that function along each line segment.
    Ok that makes sense
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  13. #13
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    Hi. Nice problem! I worked it using f(z)=-y-ix=-i\overline{z} and obtained:

    A=\sum_{j=1}^{N}(z_{n+1}-z_{n-1})\overline{z_n};\quad z_0=z_N, z_{N+1}=z_1

    However, I think a nice addition is to code it in software: Write a routine to generate a random set of complex numbers, say 10. Then find the convex hull, then calculate the area of the polygon using the build-in command for example Mathematica has build-in code to do this, then use the formula you get with the analysis above then compare the results. First do it with a square. Fun I think.
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  14. #14
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    I have a question. Do I absolutely have to use Green's Theorem or is there a way to work around this? I'm just asking because my book never really went over it. I have no problem using it, though.
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  15. #15
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    Ok I figure it all out except for the last part of the line integral of z congugate. I know that the integral of f(z)=lim summation(f(p_k)(x_k-x_(k-1)).
    My problem is just pitting this together since it starts at z_1 and not at z_0. My intial thought for f(p_k) is to be zcongugate_k. Any thoughts?
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