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Complex roots

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    find all complex roots of z^8=81i

    2. Relevant equations



    3. The attempt at a solution

    let the angle=x

    z^8=r^8(cis8x)

    we know

    81i=81 (cis pi/2)

    threfore

    z^8=81(cos pi/2 + i sin (pi/2) )

    8x= pi/2 + 2kpi
    x = pi/16 + kpi/4 kEz

    therefore

    if k=1 z=sqrt3 (cis pi/16)

    i go through this solutiosn...and should end up getting a result angle of pi/2 in one of them..becuse on of the roots is 1...so there i must be wrong..help
     
  2. jcsd
  3. Nov 26, 2007 #2

    Gib Z

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    Why must one of the solutions be 1? 81i is not equal to 1.
     
  4. Nov 26, 2007 #3

    CompuChip

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    Why do you think you can write z^8 = r^8 (cos 8x) (at least I assume that by cis you mean cos :smile:)?
    You can certainly write it as r^8 e^{8 i x} for some angle x. Also, 81i can be written in such a way (you wrote it as 81(cos pi/2 + i sin pi/2) but I might as well write it as 81 e^(i pi/2)). Now r = (81)^(1/8). What equation(s) do you get for the angle x?
     
    Last edited: Nov 26, 2007
  5. Nov 26, 2007 #4

    Gib Z

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    cis (x) is a notation engineers have adapted, an acronym for cos x + i sin x, which is equal to e^(ix).
     
  6. Nov 26, 2007 #5

    CompuChip

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    Ah, I see. Strange people, those engineers :smile:
    Anyway, your answer looks correct (at least, the one you gave, so I assume you have found the other 7 as well). Why did you think 1 was a solution?
     
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