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Thread: Complex plane area problem

  1. #1
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    Complex plane area problem

    A friend gave me this question as part of a puzzle but I am unfamiliar with how to approach. Help? Thanks

    In the complex plane, z, z^2, z^3 form, in some order, three vertices of a non-degenerate square. Determine, in lowest terms, the harmonic mean, of the 3 possible areas of these squares.
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  2. #2
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    Re: Complex plane area problem

    I might be missing something but it seems like every square, other than one of zero area, that meets these requirements will have vertices at

    $i, -1, -i, 1$

    $\large z = e^{i \pi \frac{2k+1}{2}},~k\in \mathbb{Z}$

    all these squares have sides of length $s=\sqrt{2}$ and thus area of $A=2$

    The harmonic mean is thus of course 2
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  3. #3
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    Re: Complex plane area problem

    it could also have vertices at -1+i, -2i,3-i,2+2i
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  4. #4
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    Re: Complex plane area problem

    how did you come up with that?
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  5. #5
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    Re: Complex plane area problem

    The three cases mentioned correspond to whether the diagonal of the square is the segment joining

    case 1. z to z^2

    case 2. z to z^3

    or

    case 3. z^2 to z^3

    in case 1, the line (z,z^3) must be perpendicular to the line (z^2,z^3)
    which leads to the quadratic equation

    z^3-z^2=i(z^3-z)

    similarly for the other cases

    we get three possible values for the area: 5/8, 2, 10
    Thanks from romsek
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  6. #6
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    Re: Complex plane area problem

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  7. #7
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    Re: Complex plane area problem

    Thanks for your help. I get a harmonic mean of 15/11.
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  8. #8
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    Re: Complex plane area problem

    I don't think 15/11 is the right answer. I get \frac{50}{39}

    There are 3 possible areas but there actually are 5 squares,
    two with area 10
    one with area 2
    two with area 5/8

    We should be taking the harmonic mean of the list \{10,10,2,5/8,5/8\}

    Use this fact from complex numbers:

    u,v are orthogonal and have the same magnitude if and only if u=\pm  i v
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  9. #9
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    Re: Complex plane area problem

    Based on the way the original question is stated, I think its just the three areas, not the areas of the five squares.
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  10. #10
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    Re: Complex plane area problem

    Sorry, you are right, it is 15/11
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