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Homework Help: Complex root finding

  1. Jan 9, 2009 #1
    1. The problem statement, all variables and given/known data

    let the complex number 1 = 1 + 0i. Show this number has three cube roots. Use any means to find them.


    2. Relevant equations
    not sure. cubic root x means there is some number y, x = y*y*y.


    3. The attempt at a solution

    Well one root has to be 1
    Another is (-1)*i^2 and another 1*i^2. I am not sure about these two, and I am not sure I could give an explanation why even if they are right.
     
  2. jcsd
  3. Jan 9, 2009 #2

    Dick

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    Those aren't right. Do you know deMoivre's formula? If you don't want to use that you are trying to solve x^3-1=0 for x. Factor it.
     
  4. Jan 9, 2009 #3

    HallsofIvy

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    i2= -1 so (-1)i2= (-1)(-1)= 1. That's not a new root. 1(i2)= -1 and (-1)3= -1, not 1 so that is NOT a root.

    If you are expected to do a problem like this then you should already know "DeMoivre's formula": if [itex]z= r(cos(\theta)+ i sin(\theta))[/itex] then [itex]z^c= r^c(cos(\theta)+ i sin(\theta))[/itex] which is true for c any number. In particular [itex]z^{1/n}= r^{1/n}(cos(\theta/n)+ i sin(\theta/n))[/itex] where [itex]r^{1/n}[/itex] is the unique real positive nth root of r. Since cosine and sine are periodic with period n, increasing [itex]\theta[/itex] by any multiple of [itex]2\pi[/itex] won't change [itex]r(cos(\theta)+ i sin(\theta))[/itex] but dividing by n will.

    In particular, [itex]1= 1+ 0i= 1(cos(0)+ i sin(0))= 1(cos(2\pi)+ i sin(\2pi))= 1(cos(4\pi)+ i sin(4\pi))[/itex]. Find the 3 third roots of 1 by dividing those angles by 3. For example, 1(cos(0/3)+ i sin(0/3))= 1 which gives the first third root of 1. Now, what are [itex]1(cos(2\pi/3)+ i sin(2\pi/3)[/itex] and [itex]1(cos(4\pi/3)+ i sin(\4pi/3))[/itex]?
     
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