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Complex question

  1. Oct 24, 2004 #1
    Im a first year physics student and am having trouble with this question on complex numbers, any help would be greatly appreciated:

    find all the roots of the following equation:

    z^4 + 1/2(1 - i3^(1/2)) = 0

    I know that z can be expressed as an exponential, but I don't know how or even if it helps. I tried doing it with normal algebra but you get to a stage where you have to find the 4th root of i.
     
  2. jcsd
  3. Oct 24, 2004 #2
    just read the post above this, sorry im in the wrong place
     
  4. Oct 24, 2004 #3

    HallsofIvy

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    z4- 1/2(1 - i3^(1/2)) =
    [tex]z^4- \frac{1}{2}(1- \sqrt{3}i= 0[/tex]
    is the same as
    [tex]z^4= \frac{1- \sqrt{3}i}{2}][/tex]

    The best way to solve that is to convert to "polar form":
    [tex]\frac{1}{2}- \frac{\sqrt{3}}{2}[/tex]
    becomes
    r= 1, θ= π/3 or
    [tex]cos(\frac{\pi}{3})+ i sin(\frac{\pi}{3})[/tex]
    or
    [tex]e^{\frac{\pi}{3}}[/tex]

    the fourth root of that has r= 11/4= 1 and &theta= π/12:
    [tex]cos(\frac{\pi}{12}+ i sin(\frac{\pi}{12}[/tex]
    or
    [tex]e^{\frac{\pi}{12}}[/tex]
    which, according to my calculator is abpout 0.9659+ 0.2588i.
     
    Last edited: Oct 24, 2004
  5. Oct 24, 2004 #4
    The cube root of 1 is [tex]\omega=\frac{-1+\sqrt(-3)}{2} [/tex] which is what we get when we transpose it opposite z^4.
    However, [tex]\omega =\omega^4. [/tex] So we need only look at [tex]\omega[/tex] times 1, -1, i, -i.

    In fact, it can be seen by raising to the fourth power that [tex]\omega^4-\omega=0.[/tex]
     
    Last edited: Oct 24, 2004
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