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Thread: Complex Numbers II

  1. #1
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    Angry Complex Numbers II

    Thanks for the help on my previous question, but I am struggling to find a way to solve the following problem:

    Prove that the area A(a,b,c) of the triangle in the complex plane with corners at a,b,c EC, ordered in anti-clockwise fashion, is given by the formula:

    A(a,b,c)= i/4*(a*b'-a'*b+b*c'-b'*c+c*a'-c'*a).

    Many thanks to anyone who is able to help.
    Last edited by smiler; Dec 20th 2007 at 03:02 PM. Reason: error in notation
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  2. #2
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    Hello, smiler!

    Could you clarify your notation?


    Prove that the area $\displaystyle A(a,b,c)$ of the triangle in the complex plane
    with corners at $\displaystyle a,b,c \in C$, ordered in anti-clockwise fashion, is given by the formula:

    $\displaystyle A(a,b,c) \;= \;\frac{i}{4}(aB-Ab+bC-Bc+cA-Ca)$

    What are $\displaystyle A,B,C$ ?
    . . (You've already used $\displaystyle A$ to represent the area.)


    That expression comes from: . $\displaystyle \frac{i}{4}\begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ A & B & C \end{vmatrix}$

    . . but I have no idea what I'm talking about . . .

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  3. #3
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    clarification

    sorry i forgot to include that:

    A denotes the letter a prime (a'),
    B denotes b prime (b')
    C denotes c prime (c')

    where aB implies a is multiplied by b' likewise for other components.

    thanks
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  4. #4
    Senior Member JaneBennet's Avatar
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    Do you mean complex conjugates?
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  5. #5
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    reply to jane bennett

    i am not totally sure what a' b' or c' mean but if you are wondering what a'*b means this means a' multiplied by b.

    i am not able to ask my tutor what they mean as obviously it is the christmas holiday.

    thanks
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  6. #6
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    Hello, smiler!

    I proved it . . . the looong way.


    Prove that the area of the triangle in the complex plane with vertices
    at $\displaystyle a,b,c \in C$, ordered in CCW fashion, is given by the formula:

    . . $\displaystyle A \:= \:\frac{i}{4}\left(a b'-a' b +b c' - b' c + c a' - c' a\right)$
    Theorem
    The triangle with vertices: $\displaystyle P(x_1,y_1),\,Q(x_2,y_2),\,R(x_3,y_3)$ has area given by:

    . . $\displaystyle \Delta PQR \;=\;\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{vmatrix} \;=\;\frac{1}{2}\left(x_1y_2 - x_2y_1 + x_3y_1 - x_1y_3 + x_2y_3 - x_3y_2\right)\;\;{\color{blue}[1]}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Let: .$\displaystyle \begin{array}{ccccccc}a & = & x_1+iy_2 & & a' &=&x_1 - iy_1\\ b & = & x_2 + iy_2 & & b' &=&x_2-iy_2\\ c & = & x_3+iy_3 & & c' &=&x_3-iy_3\end{array}$


    $\displaystyle ab' - a'b \;=\;(x_1+iy_1)(x_2-iy_2) - (x_1-iy_1)(x_2+iy_2) $
    . . . . . .$\displaystyle = \;x_1x_2 - ix_1y_2 + ix_2y_1 + y_1y_2 - x_1x_2 - ix_1y_2 + ix_2y_1 - y_1y_2$
    . . . . . .$\displaystyle = \;-2i(x_1y_2 - x_2y_1)$

    $\displaystyle bc' - b'c \;=\;(x_2+iy_2)(x_3-iy_3) - (x_2-iy_2)(x_3+iy_3) $
    . . . . . .$\displaystyle = \;x_2x_3 - ix_2y_3 + ix_3y_2 + y_2y_3 - x_2x_3 - ix_2y_3 + ix_3y_2 - y_2y_3$
    . . . . . .$\displaystyle = \;-2i(x_2y_3 - x_3y_2)$

    $\displaystyle ca' - c'a \;=\;(x_3+iy_3)(x_1-iy_1) - (x_3-iy_3)(x_1+iy_1)$
    . . . . . .$\displaystyle = \;x_1x_3 - ix_3y_1 + ix_1y_3 + y_1y_3 - x_1x_3 - ix_3y_1 + ix_1y_3 -y_1y_3$
    . . . . . .$\displaystyle = \;-2i(x_3y_1 - x_1y_3)$


    $\displaystyle (ab' - a'b) + (bc' - b'c) + (ca' - c'a) \;=\;-2i(x_1y_2-x_2y_1) - 2i(x_2y_3-x_3y_2) - 2i(x_3y_1 - x_1y_3)$

    . . . . . . . . . . . . . . . . . . .$\displaystyle = \;-2i(x_1y_2-x_2y_1 + x_3y_1-x_1y_3 + x_2y_3 - x_3y_2) $


    Therefore:

    $\displaystyle \frac{i}{4}(ab' - a'b + bc' - b'c + ca' - c'a) \;=\;\frac{i}{4}(-2i)(x_1y_2-x_2y_1 + x_3y_1-x_1y_3 + x_2y_3-x_3y_2)$

    . . . . . . . . . . . . . . . . . $\displaystyle = \;\frac{1}{2}(x_1y_2-x_2y_1 + x_3y_1-x_2y_3 + x_2y_3-x_3y_2) $

    which equals the formula in [1] . . . Q.E.D.

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