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Math Help - Complex cube root

  1. #1
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    Complex cube root

    Hi everyone

    got confused again.

    3+4i^(3/2) =

    THis is what i did

    z^(3/2) = 5^(3/2) = ( cos(3(2pik + 0.9273))/2 + isin3((2pik + 0.9273))/2 )

    Please advise, a bit confuse with the the ^3/2 root. should we times the whole (2pik + titah)/2 equation with 3/2 ?

    For k=0 i get
    =11.17(cos1.39095 + isin1.39095)
    k = 1
    =11.17(cos10.8157 + isin10.8157)
    k=2
    =11.17(cos20.24 = isin 20.24)

    Pls comment, im revising for my final.thank you & regards. so sorry , i got confused with the ^3/2 substitution.

    thank you & regards
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  2. #2
    MHF Contributor

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    Quote Originally Posted by anderson View Post
    Hi everyone

    got confused again.

    3+4i^(3/2) =
    This is not a "cube root"

    THis is what i did

    z^(3/2) = 5^(3/2) = ( cos(3(2pik + 0.9273))/2 + isin3((2pik + 0.9273))/2 )
    You mean 5^(3/2) times that not "=".

    Please advise, a bit confuse with the the ^3/2 root. should we times the whole (2pik + titah)/2 equation with 3/2 ?
    You should multiply by 3. That "/2" is the denominator, isn't it? Multiply (2\pi k+ 0.9273) by 3/2 which is the same, of course, as "multiply by 3 and divide by 2".

    For k=0 i get
    =11.17(cos1.39095 + isin1.39095)[/quote]
    5^{3/2}, to two decimal places, is 11.18, not 11.17.

    k = 1
    =11.17(cos10.8157 + isin10.8157)
    k=2
    =11.17(cos20.24 = isin 20.24)

    Pls comment, im revising for my final.thank you & regards. so sorry , i got confused with the ^3/2 substitution.

    thank you & regards
    Those numbers are right.
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