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Complex number equation

  1. Nov 13, 2008 #1
    Hi there, I ve got problem with this equation.

    x^4+ 14 = 0

    I tried like this:

    X^2 = z

    z^2 +14 = 0
    z^2 = -14
    z= sqr-14
    z= j 3.74

    then back to x

    x^2 = j.374

    and now what can i do??
    Last edited by a moderator: Nov 13, 2008
  2. jcsd
  3. Nov 13, 2008 #2
    Consider using the roots of unity.


    [tex] x^4 - 14 = 0 [/tex]

    [tex] x = e^\frac{2 \pi i k}{4} \sqrt[4]{14} ; k = 0, 1, 2, 3 [/tex]

    That's from memory. You'll need to double check it.
  4. Nov 13, 2008 #3


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    Both are correct, these answers are the same.
  5. Nov 13, 2008 #4


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    Messed up again! greg997, I meant to click on quote but accidently clicked on "edit".
    I hope I have re-established what you had originally.

    You should have z=-j sqrt(14).

    The two sqrts of j are [itex]\sqrt{2}/2+ j\sqrt{2}/2[/itex] and [itex]-\sqrt{2}- j\sqrt{2}/2[/itex] and the two sqrts of -j are [itex]-\sqrt{2}/2+ j\sqrt{2}/2[/itex] and [itex]\sqsrt{2}/2- j\sqrt{2}/2[/itex]. Multiply those by the fourth root of 14.

    How did I get those roots? Well, in the "complex plane", j is at (0,1). The square root will have the same modulus (1) and half the argument: 90 degrees becomes 45 degrees.
  6. Nov 13, 2008 #5
    You can do like this:

    In your case z=-14, so -14=-14+0*i

    a=-14 ; b=0

    Now [itex]r=\sqrt{(-14)^2+0^2}=14[/itex]



    At this point you need to find the angle [itex]\alpha[/itex] and substitute in the formula below.

    [tex]w_k=\sqrt[n]{r}(cos\frac{\alpha+2k\pi}{n}+isin\frac{\alpha+2k\pi}{n})[/tex] for k=0,1,2,...,n-1
  7. Nov 13, 2008 #6
    Thanks everybody for these explanations. They really helped me.
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