1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook