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Homework Help: Finding the solutions of a complex number

  1. Jan 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the 3 solutions of ei[tex]\pi[/tex]/3z3=1/(1+i)

    2. Relevant equations

    3. The attempt at a solution

    i have put i/(1+i) into polar form,1/[tex]\sqrt{2}[/tex] ei[tex]\stackrel{\pi}{4}[/tex]

    So i get z3 = [tex]\stackrel{1}{\sqrt{2}}[/tex]ei-[tex]\pi/12[/tex]

    Then i got stuck... z3=r3ei[tex]\theta[/tex]

    So shouldn't r3=1/[tex]\sqrt{2}[/tex] and [tex]\theta[/tex]=-[tex]\pi[/tex]/36 ..but that's only 1 solution?
  2. jcsd
  3. Jan 8, 2010 #2
    if [itex]z^3=\frac{1}{\sqrt{2}} e^{-\frac{i \pi}{12}}[/itex]
    then [itex]z=(\frac{1}{\sqrt{2}})^{\frac{1}{3}} e^{-\frac{i \pi}{36}}[/itex]
  4. Jan 8, 2010 #3
    but that is only one solution and it asks for 3
  5. Jan 8, 2010 #4
    You can add any multiple of 2 pi i to theta.
  6. Jan 8, 2010 #5
    ahh, of course. thank you!
  7. Jan 8, 2010 #6
    Note that for unique solutions, you need to add [tex]n\cdot 2\pi i[/tex] to the exponent of the complex number describing [tex]z^3[/tex]

    Otherwise you're just describing the same number over and over!
    Last edited: Jan 8, 2010
  8. Jan 8, 2010 #7
    haha, oh yea.
    where do you get n2/pi i from?
  9. Jan 8, 2010 #8
    That's how much you need to add to the angle of the exponent, [tex]r e^{i\theta}[/tex] so that you get the same value.


    You can easily see this using Euler's identity since the sine and cosine both have a period of [tex]2\pi[/tex] radians.

    When you take the cube root, you use De-Moivre and divide the angle by 3. Note that you get different angles depending on whether you add [tex]2\pi[/tex] once, twice, or three times to the original exponent's angle.
  10. Jan 9, 2010 #9
    yes i should have mentioned that. sorry.
  11. Jan 9, 2010 #10
    so the final answer is [tex]\stackrel{n2\pi}{3}[/tex]? i think i just misread your post as [tex]\stackrel{2n}{\pi}[/tex] or something :)
  12. Jan 10, 2010 #11
    well, is [itex](\frac{2n \pi}{3})^3=\frac{1}{\sqrt{2}}e^{-\frac{1 \pi}{12}}[/itex]?

    go to my first line of working in post 2, the other two solutions will correspond to [itex]e^{-\frac{25 i \pi}{12}}[/itex] and [itex]e^{-\frac{49 i \pi}{12}}[/itex].
    then of course, you have to do the division by 3 etc as before to get to the final answer.
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