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Cube roots

  1. Feb 8, 2006 #1
    I am asked to use the exponential form [tex]e^{i \theta}[/tex] to express the three cube roots of:

    (a) 1
    (b) i
    (c) -i

    what exactly does this question mean? I am really lost as to what they are asking for.

    here is a stab at it:
    (a)

    cube root of 1 is 1... so... would that mean.... [tex]1=e^{- \infty +i \theta[/tex]

    (b)

    cube root of i is [tex]\frac{\sqrt{3}}{2}+0.5i[/tex] so.... [tex]\frac{\sqrt{3}}{2}+0.5i=e^{i \frac{\pi}{6}}[/tex]

    (c)

    cube root of -i is [tex]\frac{\sqrt{3}}{2}-0.5i[/tex] so.... [tex]\frac{\sqrt{3}}{2}-0.5i=e^{-i \frac{\pi}{6}}[/tex]

    is this the right way to approch this problem?
     
  2. jcsd
  3. Feb 8, 2006 #2

    Tide

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    HINT: [itex]1 = e^{2\pi i n}[/itex] where n is an integer. Therefore, [itex]1^{1/3} = [/itex] ??? :)
     
  4. Feb 8, 2006 #3
    [itex]1^{1/3} =1= e^{2\pi i}[/itex] where n=1?
     
  5. Feb 8, 2006 #4

    benorin

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    [tex]1 = e^{2\pi i n}[/tex] for n=0,1,2,...
    [tex]1^{\frac{1}{3}} = \left( e^{2\pi i n}\right) ^{\frac{1}{3}}=e^{\frac{2\pi i n}{3}} [/tex] for n=0,1,2,...

    and since there are only 3 unique cube roots of 1, just put n=0,1,2.

    So [tex]1^{\frac{1}{3}} =e^{0}, e^{\frac{2\pi i }{3}},e^{\frac{4\pi i }{3}} [/tex]
     
  6. Feb 8, 2006 #5
    did I do (b) and (c) correct?

    [tex]\frac{\sqrt{3}}{2}+0.5i=e^{i \frac{\pi}{6}}=e^{i \frac{2 \pi}{6}}=e^{i \frac{4\pi}{6}}[/tex]


    something like that?
     
  7. Feb 8, 2006 #6
    for b, I got a slightly different answer.
    [tex]z^3 = i[/tex]
    [tex] \therefore z = -sin(\frac{2n\pi}{3}) + icos(\frac{2n\pi}{3}) [/tex]
    Where n = 0,1 or 2. Put the values of n in, and you will get an answer very close to what you got, but, that's assuming that I'm right (which isn't always the case!)

    You can always check what you get by cubing the complex numbers that you think the answer is. In the case of b, it should give you i.
     
    Last edited by a moderator: Feb 8, 2006
  8. Feb 8, 2006 #7
    [tex]-sin(\frac{2n\pi}{3}) + icos(\frac{2n\pi}{3})=- (i^{1/3}) [/tex]

    isnt it?
     
  9. Feb 8, 2006 #8

    Tide

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    Just for kicks, you might try setting

    [tex]i^{1/3} = x + i y[/tex]

    then cube both sides and solve for x and y. :)
     
  10. Feb 8, 2006 #9
    for (b), when I cube [tex]e^{i \frac{\pi}{6}}[/tex] i get [tex]i[/tex]
    for c, when I cube [tex]e^{-i \frac{\pi}{6}}[/tex] i get [tex]-i[/tex]

    I tried to cube x+iy but sadly, I dont know how to multiply [tex](x^2+2ixy-y^2)(x+iy)[/tex]
     
  11. Feb 8, 2006 #10

    HallsofIvy

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    Do you know DeMoivre's formula? You certainly should if you are expected to do a problem like this.
     
  12. Feb 8, 2006 #11
    nope, never heard of it. this is a challenge my teacher gave us, that's all
     
  13. Feb 9, 2006 #12

    HallsofIvy

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    DeMoivre's formula says that simply that [itex](re^{i\theta})^n = r^n e^{ni\theta}[/itex] which is obvious in exponential form! If n is a fraction, such as 1/3 then rn is the principle root.
    1 can be written as [itex]1e^{0i}[/itex] or [itex]1e^{2\pi i}[/itex] or [itex]1e^{4\pi i}[/itex]. Those clearly give the same thing but multiplying those exponents by 1/3 does not!
     
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