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Complex numbers, i don't kno if it's before or after calculus sry

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    find the three roots of z[tex]^{3}[/tex] = 1, giving non real roots in the form of e[tex]^{i\theta}[/tex], where -[tex]\pi[/tex]<[tex]\theta[/tex][tex]\leq[/tex][tex]\pi[/tex]

    2. Relevant equations

    z = re[tex]^{i\theta}[/tex]

    3. The attempt at a solution

    i kno that one of the roots is 1, an the other two form a conjugate pair. i can't find them.
  2. jcsd
  3. Nov 19, 2007 #2
    google "roots of unity" and read about how you can locate them. They are spaced evenly on the unit circle.
  4. Nov 19, 2007 #3
    and the question continues, i can't find that too.

    given that [tex]\omega[/tex] is one of the non real roots, show that 1 + [tex]\omega[/tex] + [tex]\omega^{2}[/tex] = 0
  5. Nov 19, 2007 #4
    thanks futurebird
  6. Nov 19, 2007 #5
    In general from de moivre’s theorem, when you consider a complex number in polar form you get

    z^(1/n) = r(cosx + isinx)^1/n = r^(1/n)(cos([x+2kpi]/n) + I*sin([x+2kpi]/n)) for 0<k<n-1 where k is an integer
  7. Nov 20, 2007 #6
    the thing is that i'm not familiar with de moivre's theorem. i guess we did only basic complex numbers at school. i came across this question on a past paper from another examination board. it's prolly not in our syllabus.

  8. Nov 20, 2007 #7
    I assume you know what this equation means, i.e. that this is a possible way to express a complex number (and that r>=0). Do you also know how to multiply two complex numbers?
    [tex] z_1 \cdot z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}[/tex], where [tex] e^{i2\pi} = 1[/tex] might be needed, in case you didn't already know that one.

    You are looking for z such that z*z*z = 1. Write that out in the form given and try to find solutions for z.
  9. Nov 20, 2007 #8


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    Homework Helper

    Well if you knew that z=1 was a solution to the equation [itex]z^3=1 => z^3-1=0[/itex] would that not mean that [itex](z-1)[/itex] is a factor of [itex]z^3-1=0[/itex] ?

    then you can just divide it out and you'll get [itex](z-1)(z^2+bz+1)=0[/itex] and then find the value of b...and then you use the quad. eq'n formula on the 2nd factor and get your roots...then put those roots into the form [itex]re^{i\theta}[/itex]

    but it would be easier to use the roots of unity method
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