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    Vi Nguyen's Avatar
    Vi Nguyen Posts: 48, Reputation: 2
    Junior Member
     
    #1

    Jul 2, 2009, 09:11 AM
    complex no.s
    Can anyone help me with finding the solutions to:

    (z^8)-256=0

    I can find z= 2, -2, 2i, -2i but don't know how to find the other 4 solutions, apparently I can let z^2=(a+ib)^2 and solve for a and b, but don't know how to do this. Thanks in advance.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member
     
    #2

    Jul 2, 2009, 10:34 AM
    There are n different nth roots of

    Therefore, these are given by:



    Where and n=8, so we have



    and let k=1,2,3...

    If k=1, we get 2i

    If k=2, we get

    and so on.
    Vi Nguyen's Avatar
    Vi Nguyen Posts: 48, Reputation: 2
    Junior Member
     
    #3

    Jul 5, 2009, 05:02 AM
    Hey thanks I get confused about which form to express it in to work out the solution, I had tried to put it in polar exponential form and didn't think on expanding etc.

    Quote Originally Posted by galactus View Post
    There are n different nth roots of

    Therefore, these are given by:



    Where and n=8, so we have



    and let k=1,2,3.....

    If k=1, we get 2i

    If k=2, we get

    and so on.

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