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Complex Number

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Express the complex number in exp. form
    [tex]-\frac{1}{2}(1+i\sqrt{3})[/tex]

    Solve the following eqn:
    [tex](w+2)^{4}=-\frac{1}{2}(1+i\sqrt{3})[/tex]



    2. Relevant equations



    3. The attempt at a solution
    [tex]e^{i\frac{\pi}{3}}[/tex]

    [tex]w+2=e^{(\frac{\frac{\pi}{3}+2k\pi}{4})i}
    =e^{(\frac{\pi}{12}+\frac{\pi}{2}k)i}
    [/tex]

    Such that k=0,1,2,3
     
  2. jcsd
  3. Feb 21, 2010 #2

    tiny-tim

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    Hi icystrike! :smile:

    Your first answer, eπi/3, would be the correct answer for +1/2 (1 + i√3).

    So, to get minus that, multiply by … ? :wink:

    For the second answer, the ekπi/2 can be simplified a little, for example by using a "±" (and of course, put the 2 on the other side).
     
  4. Feb 21, 2010 #3
    multiply by -1 . so it means that we have to keep a look out to or not multiply the ex. form by -1 right?
     
  5. Feb 21, 2010 #4

    tiny-tim

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    uhh? :confused:

    Hint: e? = -1 ? :smile:
     
  6. Feb 21, 2010 #5
    multiply by [tex]e^{i\pi}[/tex] thus combine the power by law of indices
     
  7. Feb 21, 2010 #6

    tiny-tim

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    Yup! … so instead of eiπ/3, it's … ? :smile:
     
  8. Feb 21, 2010 #7

    [tex]e^{\frac{4\pi}{3}}[/tex]

    yea?
     
  9. Feb 22, 2010 #8

    tiny-tim

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    (just got up :zzz: …)

    yea! :smile:
     
  10. Feb 22, 2010 #9
    Hi tiny-tim! Can you help me with this question?

    Explain why the equation [tex](z+2i)^{6}=z^{6}[/tex] has five roots.

    I thought it should be 6 roots?
     
  11. Feb 22, 2010 #10

    tiny-tim

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    Hi icystrike! :smile:

    erm :redface: … it's only a fifth-order equation! :rolleyes:
     
  12. Feb 22, 2010 #11
    hmm.. how do you tell? always thought that if we have power 6 , it will be 6 roots.
     
  13. Feb 22, 2010 #12

    tiny-tim

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    You don't like algebra, do you? :redface:

    Expand the LHS, and subtract the RHS … what happens? :smile:
     
  14. Feb 22, 2010 #13
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