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6 roots of -8 issue

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to find the 6 roots of -8 + 0i

    we are told the usual method of for these problems is to put the complex number into it's exponential form like so

    z = |z|exp( i(θ+2πk)) where k: [0 to n-1]

    then put it to the relevant power 'n'

    z1/n = |z|1/nexp(( i(θ+2πk)/n)

    so I proceed by taking θ to be pi...

    z1/6 = 81/6exp(( i(π+2πk)/6)

    thinking the question is easy...

    then in the solution given for the question the answer is 81/6exp(( i(-π+2πk)/6)

    I don't understand where this minus has come from - is the solution wrong or am I missing something?

    (picture of question and solution attached)
     

    Attached Files:

    Last edited: May 17, 2012
  2. jcsd
  3. May 17, 2012 #2

    phyzguy

    User Avatar
    Science Advisor

    Both expressions are the same 6 numbers, it's just that the value of k in one set is shifted by 1 with respect to the other set.
     
  4. May 17, 2012 #3
    You could always use
    $$
    w_k = r^{1/n}\left[\cos\frac{\theta + 2\pi k}{n}+i\sin\frac{\theta + 2\pi k}{n}\right]
    $$
    where ##n## is the number of roots in your case 6, ##\theta## is the angle, and ##k = 0,1,..,n-1##

    Then ##r = 8## and ##\theta = \pi## where the principle argument is define as ##\theta\in (-\pi,\pi]##
    So
    $$
    w_k = 8^{1/6}\left[\cos\frac{\pi + 2\pi k}{6}+i\sin\frac{\pi + 2\pi k}{6}\right] = 8^{1/6}\exp\left[i\frac{\pi + 2\pi k}{6}\right]
    $$

    So how could ##\pi## be negative? Simply define the the principle argument to be ##\theta\in [-\pi,\pi)##.

    And sometimes it will be defined be excluding ##\pi## all together like ##\theta\in (-\pi,\pi)##
     
  5. May 17, 2012 #4

    jedishrfu

    Staff: Mentor

    what minus sign? I see the -8+0i which you've handled by using pi for theta.

    your answer looks right as it is.
     
  6. May 17, 2012 #5
    ah i see when I take the principle arguments of each root then the answers agree - thanks all. Still don't understand why he decided to do it like that but I guess it doesn't matter :)
     
  7. May 17, 2012 #6

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    attachment.php?attachmentid=47370&d=1337277366.jpg

    You're supposed to find the 6 sixth roots of -8 .
     
  8. May 17, 2012 #7
    yeah I know that - .... e^i*pi = -1

    8 e^i*pi = -8

    e^2*pi*k =1 for k: 0,1,2,3....
     
  9. May 17, 2012 #8

    jedishrfu

    Staff: Mentor

    and when you plot them in the complex plane it will look like a six evenly spaced slices of a pizza pi with a centerline about the imaginary axis.
     
  10. May 17, 2012 #9
    And of course you can write ##\sqrt{2}## in place of ##8^{1/6}## in your answer.
     
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