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Complex Trigo Equation

  1. Apr 3, 2009 #1

    Mentallic

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    Homework Helper

    1. The problem statement, all variables and given/known data

    Solve [tex]z^4-2z^3+5z^2-2z+1=0[/tex]


    2. Relevant equations

    let [tex]z=cos\theta +isin\theta[/tex]

    [tex]z^n+z^{-n}=2cos(n\theta)[/tex]

    [tex]cos2\theta=2cos^2\theta -1[/tex]


    3. The attempt at a solution

    Since [itex]z\neq 0[/itex] dividing through by [itex]z^2[/itex] yields:

    [tex](z^2+z^{-2}) -2(z+z^{-1})+5=0[/tex]

    Thus, [tex]2cos2\theta - 4cos\theta +5=0[/tex]

    Simplified: [tex]4cos^2\theta -4cos\theta+3=0[/tex]

    This is a quadratic in [itex]cos\theta[/itex] that doesn't have any real solutions:

    [tex]cos\theta=\frac{1}{2}\left(1\pm \sqrt{2}i \right)[/tex]

    I've checked through my working thoroughly so I'm quite sure there aren't any mistakes so far, but I wouldn't know how to actually solve this equation's complex roots. I guess what I'm asking is how do I solve:

    [tex]cos^{-1}\left[ \frac{1}{2}\left(1\pm \sqrt{2}i \right) \right][/tex]
     
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  3. Apr 3, 2009 #2

    HallsofIvy

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    You know, I presume, that [itex]e^{i\theta}= cos(\theta)+ i sin(\theta)[/itex] and, since cosine is an even function and sine an odd function, that [itex]e^{-i\theta}= cos(\theta)- i sin(\theta)[/itex]. Adding those two equations, [itex]e^{i\theta}+ e^{-i\theta}= 2 cos(\theta)[/itex] and, finally, [itex]cos(\theta)= (e^{i\theta}+ e^{-i\theta})/2[/itex]

    Solve the equation [itex](e^{i\theta}+ e^{-i\theta})/2= \frac{1}{2}(1\pm i\sqrt{2})[/itex]. It would probably be best to let [itex]x= i\theta[/itex] first so that equation becomes [itex]x+ x^{-1}= 1\pm i\sqrt{2}[/itex] which reduces to a quadratic for x.
     
  4. Apr 3, 2009 #3

    Mentallic

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    Ahh thanks Hallsofivy, nice tip :smile:

    I end up with the result [tex]z=\frac{1 \pm _1 \sqrt {-3 \pm _2 4\sqrt{2}i}{2}[/tex]

    But I cannot simplify it to remove the imaginary unit from the surd. I'm sure an answer like this wouldn't be satisfactory. Maybe converting it into mod-arg form by approximation and reconverting back into [itex]a+ib[/itex] form by another approximation? of course, exact answers would be nicer :wink:
     
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