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Imaginary nos.

  1. Feb 17, 2005 #1
    Folks, I was just wondering why I can write:

    [tex]i^{4/3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}[/tex]

    Regards
     
  2. jcsd
  3. Feb 17, 2005 #2

    Curious3141

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    Do you know the identity [tex]e^{i\theta} = \cos\theta + i\sin\theta[/tex] ?

    [tex]i = e^{i\frac{\pi}{2}}[/tex]

    Use that, and the basic rules for exponentiation to get the result. Then express it in Cartesian form.
     
  4. Feb 17, 2005 #3

    dextercioby

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    Sure,complex exponentiation is multivalued,meaning that "i" can be written in an infinte ways using various arguments of the exponential...But worry about that less and use the formula which (though incomplete,hence inaccurate) was already given...

    Daniel.
     
  5. Feb 17, 2005 #4

    Curious3141

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    True, I know exponentiation is multivalued, but since he just wanted the principal value, that's what I gave him.
     
  6. Feb 17, 2005 #5

    dextercioby

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    Sorry to be nit-picking,but nowhere in his post is that "he just wanted the principal value" noticeable...

    Daniel.

    P.S.It doesn't matter,it's good if the OP got the simple part,at least.
     
  7. Feb 17, 2005 #6

    Curious3141

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    OK...

    [tex]i^{\frac{4}{3}} = e^{i(\frac{\pi}{2} + 2k\pi)(\frac{4}{3})}[/tex]

    The possible values in Cartesian coords are what was given, [tex]-\frac{1}{2} -i\frac{\sqrt{3}}{2}[/tex] and 1.

    I know this. The point is, he just wanted help to "see" the first answer.
     
  8. Feb 17, 2005 #7

    ehild

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    You can write if you add "or 1, or

    [tex]-\frac{1}{2}-i\frac{\sqrt{3}}{2}[/tex]",

    as

    [tex]i^{4/3}=(i^4)^{1/3}=\sqrt[3]{1} [/tex].

    1 has got n different n-th roots,

    [tex]\sqrt [n] {1} = \cos(\frac{2\pi}{n}k) + i \sin (\frac{2\pi}{n}k) [/tex]
    k=0, ...n-1.
     
  9. Feb 17, 2005 #8
    Thank you for all the help! I think I finally get what you're saying...

    [tex]z=i\, ^{\frac{4}{3}}=\left( i^4 \right)^{1/3} = \sqrt[3]{1}[/tex]

    [tex]h= 1 = \mbox{cis } 0[/tex]

    [tex]z=h_w ^3 = \mbox{ cis } \left( \frac{0+2k\pi}{3} \right) = \cos \left( \frac{2k\pi}{3} \right) +\mbox{ } i\mbox{ } \sin \left( \frac{2k\pi}{3} \right) = -\frac{1}{2} +\mbox{ } i \mbox{ } \frac{\sqrt{3}}{2}[/tex]
     
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