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Pentagram coordinates

  1. Jan 23, 2005 #1
    Hi,
    I tried to find out how to plot a pentagram in the complex plane. Let the radius of the pentagram be 1, then all the corners of the pentagram satisfy the equation
    x^5 = 1 (where x is a complex number).

    The answer is, of course, x = exp (2n*pi*i/5).

    But I wanted to express that in roots, not in exponentials.
    I could not find anything on the Web, so I tried it myself.
    It can be done, but the answer is not beautiful.
    The prime root is:
    x = a + i*a / sqrt(1-2/sqrt(5)),
    where a = [ (9-4*sqrt(5)) / (16*sqrt(5)-16) ]^(1/5).
    Next, I tried to express this in terms of the "Golden Ratio" Phi. Because I knew that Phi appears in the pentagram a lot.
    Since
    Phi^2 = 1*Phi + 1
    Phi^3 = 2*Phi + 1
    Phi^4 = 3*Phi + 2
    Phi^5 = 5*Phi + 3
    ... and so on (that's the Fibonacci series twice),
    we get
    x = (Phi - 1)/2 + sqrt(Phi + 2)*i/2
    and also
    x^2 = -Phi/2 + sqrt(-Phi + 3)*i/2.
    (x^3 and x^4 are obviously symmetrical wrt. the real axis)
    That looks better.

    But not perfect. I tried to get rid of the square root in the imaginary part, because I wanted everything to be linear in Phi. But no success.

    Any help?
     
    Last edited: Jan 23, 2005
  2. jcsd
  3. Jan 23, 2005 #2

    Hurkyl

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    I'm pretty sure that one coordinate will require a square root nested inside a square root.
     
  4. Jan 23, 2005 #3

    dextercioby

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    Well,they do.Using
    [tex] \varphi=:\frac{1+\sqrt{5}}{2} [/tex]
    ,u see that sqrt of \phi implies sqrt from an sqrt...

    Daniel.
     
  5. Jan 23, 2005 #4

    Hurkyl

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    Right -- I was suggesting that he's stuck with something involving the square root of phi.

    IMHO, it's better to explicitly write out what it is, rather than suppressing the square root of 5 into phi.
     
  6. Jan 24, 2005 #5
    Thanks Hurkyl.
    Of course, in some cases, a root containing Phi can be removed. For instance,
    sqrt(Phi + 1) = phi
    sqrt(3*Phi + 2) = phi + 1,
    etc.

    Do you think it's impossible for
    sqrt(Phi + 2) = ?
    sqrt(-Phi + 3) = ?
     
  7. Jan 24, 2005 #6

    Hurkyl

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    Suppose it can be written as a * phi + b. Now square both sides and try to solve for a and b...
     
  8. Jan 25, 2005 #7
    Yes. It works, but the coefficients have a square root inside a square root.
     
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