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Complex- area enclosed by a polygon

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,zn is

    Show that the area enclosed =1/2I[tex]\Sigma[/tex]zkcongugate(zsub(k+1)-zk).
    Interpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz.

    2. Relevant equations

    3. The attempt at a solution
    I don't even know where to start. I understand where the 1/2I(z.....) comes from, but once the summation comes in, I'm lost.
  2. jcsd
  3. May 1, 2009 #2
    This really is not as hard as you think, especially since you already believe A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ]

    For the summation 1/2 [ Im( bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n) ] just "remove parentheses" and observe that bar(z) z=|z|^2 so its imaginary part is 0.

    As for the line integral, use the fact that \int_C f(z) dz is approximately \lim\sum_{k=1}^n f(z_k^*)(z_k-z_{k-1})
  4. May 4, 2009 #3
    Do i need to show using green's thm?

    I did some work and have a couple questions:
    Let x be polygon with its ordered vertices (the vertices you take in order
    as you travel around it) be .

    Also say they are taken in a counterclockwise fashion.
    Let x be the line segment from zsubj,zsub(j+1) for j=1,2,...n-1. But for
    j=n define x as[zsubn,zsubn].

    Define C-->C by f(x+iy)=-y+ix. I'm iffy on this part because I wasn't
    quite sure how to define f(x+iy).
    Let R denote the interior of the polygon, then by Green's theorem:
    1/2integral(f)=double integral(Green's Thm)=R

    Then I use a summation of integrals for each line segment.

    For j=1,2,3,,,n-1 define y=zsubj +(zsub(j+1)-zsubj)t for 0<t<1.
    Notice that y parametrizes [zj,zj+1].

    f(x+iy)==y+ix=i(x-iy) so f(z)=izconjugate while y=zj+1-zj

    Then we evealuate about the integral form 0 to 1.

    So adding all up, we get

    I got this far, but don't really understand what I'm doing. My biggest
    problem is using Green's Theorem and understanding my choices for f(x+iy)
    and why i parameterized.
    I'm looking for some help explaining what I did and where to go from here.
  5. May 4, 2009 #4
    I was thinking more geometrically. I didn't use Green's Theorem at all.

    A triangle with vertices (0,0), (a,b), and (c,d) has area 1/2 * (ad - bc). This is a fact from analytic geometry. (The cross product of two vectors has a magnitude equal to the area of a certain parallelogram, and this triangle has half that area.)

    So now, if you call z1 = (a,b) = a + bi and z2 = (c,d) = c + di, then it is interesting that 1/2 * conjugate(z1) * z2 has imaginary part equal to 1/2 * (ad - bc).

    Now generalize slightly, so that the third vertex is not at the origin, but instead is at z3. If you use translation, you can derive a formula for the area of this triangle.

    Now suppose you have a convex polygon with vertices z1, z2, ... , zn. Draw segments from z1 to each of the other vertices, dividing the polygon into triangles. Add up the areas.

    This is how you can obtain the formula A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ].

    Maybe I'm not understanding your question.
  6. May 4, 2009 #5
    I don't think i show using geometry.
  7. May 5, 2009 #6
    I'm having trouble going from a triangle from a vertice of (0,0) to one without. My probelm is finding the area. Also I'm confused on going from our area formula to the summation. Any hints?
  8. May 5, 2009 #7
    Like I see how to show with the 3 vertices with one being (0,0) but then when I try to go to the summation, I'm not sure how to show that. Can someone help me show that? I know once I get there I will be able to get the complex line integral.
  9. May 5, 2009 #8
    Ok, I figured out how to show that the summation area was the same with the (0,0) vertice by plugging in the vertices points. So I figured that out.
  10. May 6, 2009 #9
    Ok, I'm all set except fo using a line integral about z congugate using the definition of a summation. i know the line integral of f(x) =Lim summation[f(psubk)(xsubk-xsubk-1)]
    My problem is seeting this up. My initial assumption is to do (zsubk congugate)z_kcongugate -z-(k-1))? I think if I were to be able to set it up right, i'd be fine on evaluating it.
  11. May 6, 2009 #10
    Well, can't you say for a simple closed path C, that [tex]\int_C bar(z)\,dz[/tex] is approximated like this: take points [tex]z_1, z_2, \dots, z_n[/tex] around the path C and on each segment [tex][z_i,z_{i+1}][/tex] let [tex]\zeta_i = z_i[/tex], so that the integral is approximately

    [tex]bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)[/tex]

    But the imaginary part of this sum is twice the area bounded by the polygon with vertices [tex]z_1, z_2, \dots, z_n[/tex]

    Then say something about the limit.
  12. May 7, 2009 #11
    So take the limit of bar zk(z_k+1-z_k)
    which will end up being 1/2[zk+z_k+1](z_k+1-zk}
  13. May 7, 2009 #12
    I'm sorry, I don't follow what you are saying when you say "which will end up being 1/2[zk+z_k+1](z_k+1-zk}"

    In the "Riemann sum," you don't have to take the midpoint of segment [tex][z_i,z_{i+1}][/tex]

    You can take any point on the segment. I choose to take [tex]z_i[/tex]

    However, I don't know if this comment is related to what you are saying or not.
  14. May 7, 2009 #13
    Ok then i'm a bit confused since i want to evaluate using the limit definition and want to show an answer like how if I chose to just do zdz I would get 1/2(b^2-a^2). Do you see what I'm seeing?
  15. May 7, 2009 #14
    Maybe I see what you are meaning.

    But I don't mean evaluate a limit. The original question was to "Interpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz."

    So as I understand the problem, you simply make the statement or observation that [tex]\int_C bar(z)\,dz[/tex] is approximately [tex]bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)[/tex]
  16. May 7, 2009 #15
    I see what you mean. Ok then, what if I wanted to go a little further and evaluate using that limit definition.
  17. May 7, 2009 #16
    Oh, so it has to be half the sum of the imaginary parts because of the imaginary parts being twice the area?
  18. May 7, 2009 #17
    And I can say something about when evaluating the limit we use 1/2 of the imaginary part?
  19. May 7, 2009 #18
    [tex]bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)\approx\int_C bar(z)\,dz[/tex]

    so take imaginary parts of both sides and divide by 2
  20. May 7, 2009 #19
    Ok, I think this makes sense because we are picking a point in the interval and from that we get that, but we get twice the area, so we must divide by 2. Correct me if I'm wrong.
  21. May 8, 2009 #20
    This is how I analyzed it (hope I don't confuse things):

    We can start with the known relation for the area encircled by a closed contour (see Calculus text book, "Topics in Vector Calculus") :

    [tex]\frac{1}{2}\mathop\oint\limits_{C} (-ydx+xdy)=\mathop\iint\limits_{R} dA[/tex]

    Note the expression [tex]-ydx+xdy[/tex] is the real part of the contour integral:

    [tex]\oint f(z)dz=\oint udx-vdy+i\oint vdx+udy[/tex]

    with [tex]f(z)=-y-ix=-i\overline{z}[/tex]

    Now, by Green's Theorem:

    \oint vdx+udy=\oint (-xdx-ydy)=\mathop\iint\limits_{R}\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0

    and therefore:

    [tex]\mathop\iint\limits_R dA=-\frac{i}{2}\oint \overline{z}dz[/tex]

    We can parameterize the straight-line segments [tex](z_n,z_{n+1})[/tex] as:

    and therefore:

    \mathop\oint\limits_{z_n(t)} \overline{z}dz&=\int_0^1 \overline{z_n(t)}(z_{n+1}-z_n)dt \\
    &=(z_{n+1}-z_n)\int_0^1 \left(\overline{z_n}+\overline{(z_{n+1}-z_n)}t\right)dt\\

    \mathop\iint\limits_{R} dA&=-\frac{i}{4}\sum_{j=1}^N (z_{n+1}-z_n)\left(\overline{z_n}+\overline{z_{n+1}}\right);\quad z_{N+1}=z_1 \\
    &=-\frac{i}{4}\sum_{j=1}^N\big(z_{n+1}-z_{n-1}\big)\overline{z_n};\quad z_0=z_N,\:z_{N+1}=z_1
    Last edited: May 8, 2009
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