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Complex algebra problem (roots)

  1. Aug 12, 2015 #1
    1. The problem statement, all variables and given/known data
    First off i wasn't sure if i should put this in precalc or here so i just tossed a coin

    I must find the roots of the expression z^4 +4=0 (which i've seen repeatedly on the internet)
    Use it to factorize z^4 +4 into quadratic factors with real coefficients
    The answer is (z^2 + 2z + 2)(z^2 - 2z + 2)

    2. Relevant equations
    Just the one given on the top

    3. The attempt at a solution
    Ok so, the question is quite basic as i've seen people pull it off in 10 mins, BUT after 3 semesters of dealing with only real numbers i forgot about complex ones, and now that i took complex analysis i'm in trouble D:
    First off i had to search Euler's identity, that's how lost i was
    I managed to establish that the magnitude of Z (r) is sqrt(2), i also got to the point where phi is -pi/4 +k*pi/2
    so the 4 roots would be given by the sqrt(2) * e^i*phi given k takes the values within (0,1,2,3)
    This is where i die i have no idea how to turn those into the final equation, i know the 4 solutions are
    (+/-)1 (+/-)i, and from there is just polynomial algebra, but i don't know how to get there, Am i supposed to be able to deduct this as obvious from my last step?
    Thanks in advance, i firmly believe there are no such thing as dumb questions, but given this is something I'm supposed to manage, it's quite the candidate :D
  2. jcsd
  3. Aug 12, 2015 #2

    Ray Vickson

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    You can write ##-1## as
    [tex] -1 = e^{i \pi} = e^{3 i \pi} = e^{-i \pi} = e^{-3 i \pi},[/tex]
    so the four 4th roots of ##-1## are obtained by taking 1/4 of the argument of each of the above. That is, the four roots ##r_1,r_2,r_3,r_4## are
    r_1 = e^{i \pi/4} = (1+i)/ \sqrt{2}, \; r_2 = e^{i 3 \pi/4} = (-1 + i)/ \sqrt{2} \\
    r_3 = e^{-i \pi/4} = (1-i)/ \sqrt{2}, \; r_4 = e^{-i 3 \pi/4} = (-1-i)/ \sqrt{2} [/tex]
    From these you can determine the four 4th roots of ##-4##.

    Note that ##e^{i \theta} = \cos(\theta) + i \sin(\theta)##, so we can easily write the four roots in ##a + ib## form, just using elementary geometry/trigonometry.
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