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How to determine how many distinct members of set

  1. Jan 12, 2008 #1
    let
    A = {z(belogs to)C | z^6 = −64}
    list all members of A
    how many distinct members of A are there?
     
  2. jcsd
  3. Jan 12, 2008 #2
    Write [tex]-64=2^6(\cos(2\,k+1)\,\pi+i\,\sin(2\,k+1)\,\pi)[/tex], thus

    [tex]z^6=-64\Rightarrow z=2\,\left(\cos\frac{(2\,k+1)\,\pi}{6}+i\,\sin\frac{(2\,k+1)\,\pi}{6}\right)[/tex]

    Now it is easy to count the distinct members of A.
     
  4. Jan 12, 2008 #3

    HallsofIvy

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    The equation zn= a where a is a complex number, always has n distinct solutions in the complex numbers. They are equally spaced around a circle centered on 0 with radius the positive nth root of |a|.
     
    Last edited: Jan 12, 2008
  5. Jan 12, 2008 #4
    how did you get 2k+1? and how is it easy to count the distinct members?
    i'm sorry i know it should be easy but i cant figure out the answer.
     
  6. Jan 12, 2008 #5
    The [itex]2\,k+1[/itex] is used in order to write the -1, with trigonometric functions, i.e.

    [tex]\cos(2\,k+1)\,\pi=-1,\quad \sin(2\,k+1)\,\pi=0[/tex]

    As HallsofIvy posted there are 6 distinct solutions. If you plug in the equation

    [tex]z=2\,\left(\cos\frac{(2\,k+1)\,\pi}{6}+i\,\sin\frac{(2\,k+1)\,\pi}{6}\right)[/tex]

    the values [itex]k=0,1,2,3,4,5,\dots[/itex] then you will see that after the 6th value the solutions, repeat themselves.
     
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