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Finding the nth roots of a complex number!

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    find the sixth roots of i.



    2. Relevant equations



    3. The attempt at a solution

    So I started by

    Arg(z)=pi/2 and |z|=1=r n=6


    so z= r^(1/6)*e^i((5kpi)/12) for k=0,1,2....n-1

    and thats as far as I got

    and there answer = e^i*n*pi/12 for n= -1, -7, -3, 1, 5, 9
     
  2. jcsd
  3. Apr 12, 2012 #2
    Did you add ([itex]\frac{\pi}{12}[/itex] + [itex]\frac{4k\pi}{12}[/itex]) to get [itex]\frac{5k\pi}{12}[/itex]?

    I don't think you can just add it like that, k is a variable.

    It would be like adding 3 + 3x and saying that's 6x.

    Leave it in the expanded form, and then sub in your values of k, that should give you matching answers to the solution, or so I think, check it again I didn't write it out and try it.
     
  4. Apr 12, 2012 #3
    Ok that makes sense but I run into problems when trying to find the negative values ie, I get

    1*e^(n*pi)/12 for n=1,5,9.... then I get 13,17,21 but the answer says -11,-7,-3... I can just work it out since I realised the roots are in = increments so i can subtract 4 from 1 then subtract 4 from -3 and so on....

    but I was curious as to why I run into trouble here??
     
  5. Apr 12, 2012 #4
    I'm not too sure what you're doing with the substitutions, but it might be your approach.

    Backtrack a bit to where you calculated up to here:

    ([itex]\frac{\pi}{12}[/itex] + [itex]\frac{4k\pi}{12}[/itex])

    Instead of doing what you did by adding it. Now substitute in values for k.

    For e.g. k = 0, you will get [itex]\frac{\pi}{12}[/itex], if you take k = 3, you know that will go outside the bound of the principal argument, so you select k = 2, and you get [itex]\frac{9\pi}{12}[/itex]

    Compare these two to the answer, they match for n = 1, and n = 9 don't they?

    Essentially, when you substitute a k-value your final answer becomes something of the form [itex]\frac{n\pi}{12}[/itex]

    For now, use the method you were doing earlier, substitute k-values get your 6 roots, and then you will see how the match up with [itex]\frac{n\pi}{12}[/itex], you will get the negative values by simply substituting negative k-values.
     
  6. Apr 12, 2012 #5
    ok that was basically what I was doing it but I thought k was restricted to the positive integers.?
     
  7. Apr 12, 2012 #6
    Why would k be restricted to positive integers only? Unless the question specifically asks for the roots when k is positive, but it doesn't.
     
  8. Apr 12, 2012 #7
    The way someone showed me was that you add 2kpi to get more solutions where k=1,2,3....n-1.

    By the way can you tell me if this is right??

    Find all complex fourth roots of -1-i

    so arctan(1)=pi/4 since its in the third quad then subtract pi so Arg(z)= -3pi/4
    and |z|=sqrt(2)

    so 2^1/8((cos((-3pi/4)+2kpi)+isin((-3pi/4)+2kpi))/n)

    = 2^1/8 {(cos[-3pi+8kpi]+isin[-3pi+8kpi])/16} ----> for k =-1,0,1,2

    =2^1/8 e^i(npi)/16 where n = -11,-3,5,13

    ???
     
  9. Apr 12, 2012 #8
    Yes, those look correct, but beware of how you're expressing fractional exponents in polar form.

    This is a bit hard to read, but...

    {(cos[-3pi+8kpi]+isin[-3pi+8kpi])/16}

    That 16 should be included in the theta value for cos and sin.

    cos(-3Pi/16) is not the same thing as cos(-3Pi)/16

    Remember, you can always easily check your answers using online calculators.

    http://www.wolframalpha.com/input/?i=roots+z^4+%3D+-1+-i

    And yes, you add 2kPi to get more solutions as you've done here, but you can still use negative values of k, as that negative signs means you're going the other way.
     
  10. Apr 12, 2012 #9

    HallsofIvy

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    Your error is multiplying the exponent for the sixth root by k. You need and additive term.

    Since sine and cosine are periodic with period [itex]2\pi[/itex], the exponential is periodic with period [itex]2\pi i[/itex]. So [itex]i= e^{(\pi/2+2k\pi)i}[/itex] for any integer i.

    From that, [itex]i^{1/6}= e^{(\pi/12+ k\pi/3)i}[/itex]. k can be any integer but a given k and k+ 6 or k- 6 will give the same number so there are exactly 6 distinct sixth roots.
     
  11. Apr 12, 2012 #10
    The only problem with wolfram is I cant really seem to get it in polar form so I can check my answers.

    Although I just purchased maple so hopefully I will be able to work that out.
     
  12. Apr 12, 2012 #11
    Is this answer for the original problem or the later posted??
     
  13. Apr 12, 2012 #12
    It's for the original of course...he's talking about the 1/6 roots of i.

    Getting some of the answers in a form you can understand in MAPLE can be a bit tricky.

    Generally it's convert(%,polar) and evalc(%) that moves things around with Re(%) and Im(%)
     
  14. Apr 12, 2012 #13

    Yeah ok thanks heaps for the help its great!!
     
  15. Apr 13, 2012 #14
    Start with an easier problem. Find the square root of i. Do it by drawing i on the complex plane and visualizing geometrically why the square root of i is what it is. Generalize.

    I don't know if that's too vague but you should literally be able to do this problem in your head once you have the correct visualization.
     
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