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Math Help - Roots

  1. #1
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    Roots

    Hi

    Got problem with these questions any suggestions?
    Show that 2-i is a root of z^4 - 8z^3+28z^2-48z+35. Hence find all roots
    and
    Find all 5th roots of -32 and all 4th rots of 81i
    In form x+iy where x and y in 4 DP
    Thx
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  2. #2
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    Quote Originally Posted by NeloAngelo View Post
    Hi

    Got problem with these questions any suggestions?
    Show that 2-i is a root of z^4 - 8z^3+28z^2-48z+35. Hence find all roots
    all the coefficients of the quartic are real so the complex conjugate is a root.
    so we have two factors (x-(2-i)) and (x-(2+i)).
    so (x-(2-i))(x-(2+i)) is a factor of your quartic, divide your quartic by this factor and then solve the quadratic to get the two remaining roots.

    Find all 5th roots of -32 and all 4th rots of 81i
    you want the solve the equation z^5 = -32. first rewrite -32 in the form of Re^{i(\theta +2\pi n)}

    tell me how it goes

    Bobak
    Last edited by bobak; April 17th 2008 at 07:43 AM.
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  3. #3
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    Quote Originally Posted by bobak View Post
    all the coefficients of the quartic are real so the complex conjugate is a root.
    so we have two factors (x-(2-i)) and (x-(2+i)).
    so (x-(2-i))(x-(2+i)) is a factor of your quartic, divide your quartic by this factor and then solve the quadratic to get the two remaining roots.



    you want the solve the equation z^5 = -32. first rewrite -32 in the form of Re^{i(\theta +2\pi n)}

    tell me how it goes

    Bobak

    Thx for the feedback
    How did you get those answers on the 1st part? Did you use combination of de Moivre equation, quadratic formulae and long division
    o how would it look if I applied -32 to Re^{i(\theta +2\pi n)} ?? Just lookin at tht formula confused the hell outta me
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  4. #4
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    Quote Originally Posted by NeloAngelo View Post
    Thx for the feedback
    How did you get those answers on the 1st part? [snip]
    Two things you should know:

    1. If \alpha is a root then z - \alpha is a linear factor.

    2. If a polynomial has real coefficients and \alpha is a root, then \bar{\alpha}, the conjugate of \alpha, is also a root.

    Some of the questions you're asking suggest it would be a good thing for you to go back and thouroughly review this whole topic.
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