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Math Help - Determine Roots of Complex Polynomial

  1. #1
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    Determine Roots of Complex Polynomial

    Hello,

    I encountered this problem in my exercise set, and was stumped.

    Solve (z + 1)^4 = 1 - i

    My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.

    Which then gives..

    (z + 1)^4 = (1 - i)^4 / 8

    (z + 1) = +/- (1 - i) / 4throot(8)

    But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.

    Edit: Sorry, but LaTeX kept giving me an error when I tried using it.
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    MHF Contributor chisigma's Avatar
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    Re: Determine Roots of Complex Polynomial

    Quote Originally Posted by TaylorM0192 View Post
    Hello,

    I encountered this problem in my exercise set, and was stumped.

    Solve (z + 1)^4 = 1 - i

    My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.

    Which then gives..

    (z + 1)^4 = (1 - i)^4 / 8

    (z + 1) = +/- (1 - i) / 4throot(8)

    But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.

    Edit: Sorry, but LaTeX kept giving me an error when I tried using it.
    Setting s=z+1 Your equation become...

    s^{4}= 2^{\frac{1}{2}}\ e^{i(2k-\frac{1}{4})\pi} (1)

    ... and its solution are...

    s= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}

    ... so that the solutions of the equation in z are...

    z= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}-1 (3)

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Determine Roots of Complex Polynomial

    Hmm...I like this solution! Thank you!

    Our text stipulates, however, that we should follow the "outline" used in one of the examples; i.e. with some clever algebraic manipulation put the complex polynomial equation in a solvable Cartesian form (i.e. do not use the polar form of the complex numbers). When I attempted it this way, I got stuck as I mentioned.

    In any case, I think it's pretty dumb when in a few steps you can see (much more clearly at that) what the roots are.

    I do have one question, though...is it acceptable to leave the solution in a non-standard form? i.e. the polar form with a "-1" tacked at the end? Or should you expand the polar form into its trigonometric representation and go through the grueling task of algebraically rewriting the expression in a standard form? If it is acceptable...then is it possible in this non-standard form to plot the roots (by inspection) on the complex plane, or is this pretty much an "analytic only" exact answer?

    Thanks ~
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