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Math Help - forgot how to solve nth root

  1. #1
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    forgot how to solve nth root

    (a)
    how do you find the fifth root of -16 ( 2) ^ (1/2) ( 1+i)

    i found that the mod of z^5 is 4. and the argument of ( 1+i ) = (Pi /2)

    from there, i said that the arg of z is ( Pi / 10 + 2k Pi / 5)

    thus the answer for the fifth root of z would be of the form,

    z = 4 (cos ( Pi / 10 + 2k Pi / 5) + i sin (Pi / 10 + 2k Pi / 5) )


    am i right to say that?

    (b)
    also is there a fixed method that i can do to compute questions like ( 1-i) ^ ( -3) and ( -1 + i) ^16?

    becos now im just doing them via expansion and im wondering if there is a shorter method that i can do to solve it..

    thanks!
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    (a)
    how do you find the fifth root of -16 ( 2) ^ (1/2) ( 1+i)


    What did you write here? Is it -16\sqrt{2}(1+i)?


    i found that the mod of z^5 is 4. and the argument of ( 1+i ) = (Pi /2)


    This is wrong if the number is what I wrote above: if z=-16\sqrt{2}(1+i)=-16\sqrt{2}-16\sqrt{2}i , then |z|=\sqrt{16^2\cdot 2+16^2\cdot2}=\sqrt{4\cdot 16^2}=2\cdot 16=32
    Also, its argument is Arg(z)=\arctan\left(\frac{Im(z)}{Re(z)}\right)=\ar  ctan 1=\frac{\pi}{4}+k\pi\,,\,\,k\in\mathbb{Z}

    So try now to solve your question with the above info...and check how come you thought it was something else!

    Tonio



    from there, i said that the arg of z is ( Pi / 10 + 2k Pi / 5)

    thus the answer for the fifth root of z would be of the form,

    z = 4 (cos ( Pi / 10 + 2k Pi / 5) + i sin (Pi / 10 + 2k Pi / 5) )


    am i right to say that?

    (b)
    also is there a fixed method that i can do to compute questions like ( 1-i) ^ ( -3) and ( -1 + i) ^16?

    becos


    Oh, and please: it is "because"...common, a little university level's basic writing.



    now im just doing them via expansion and im wondering if there is a shorter method that i can do to solve it..

    thanks!
    .
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