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Thread: finding a complex numbers

  1. #1
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    finding a complex numbers

    Hi folks,

    I am asked to use the definition $\cos z = \dfrac{e^{iz} + e^{-iz}}{2}$ to find 2 imaginary numbers having a cosine of 4.

    so, $\cos z = 4$ therefore

    $4 = \dfrac{e^{iz} + e^{-iz}}{2} $

    $8e^{iz} = e^{i2z} + 1$

    let $w = e^{iz}$

    $w^2 - 8w + 1 = 0$

    $w = 4 \pm \sqrt{15}$

    $e^{iz} = 4 \pm \sqrt{15}$

    $iz \ln e = \ln (4 \pm \sqrt{15}) $

    multiply both sides by i

    $i^2 z = i \ln(4 \pm \sqrt 15)$

    $ -z = i \ln(4 \pm \sqrt 15)$

    $z = i \ln \dfrac{1}{(4 \pm \sqrt 15)}$

    the actual answer is

    $z = i \ln (4 \pm \sqrt{15}) $

    can anyone see what I am doing wrong?
    Last edited by s_ingram; May 28th 2018 at 09:33 AM.
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  2. #2
    Junior Member Walagaster's Avatar
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    Re: finding a complex numbers

    Your answers are correct and the "actual answer" is wrong.
    Last edited by Walagaster; May 28th 2018 at 10:16 AM.
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  3. #3
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    Re: finding a complex numbers

    Well, that's a first! Me being right and the book wrong! I think I will start a small celebration. Where's that bottle?
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  4. #4
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    Re: finding a complex numbers

    the book is also right
    Thanks from s_ingram
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  5. #5
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    Re: finding a complex numbers

    I did consider this possibility. I thought that it may be due to the plus or minus sign. It could absorb the -z, but I wasn't convinced and sought the wisdom of the experts. I must admit Idea, that I like your minimalist style, but this time, it is too minimal. Could you kindly explain?
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  6. #6
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    Re: finding a complex numbers

    $\displaystyle \left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=1$

    therefore

    $\displaystyle \ln \left(4+\sqrt{15}\right)=\ln \left(\frac{1}{4-\sqrt{15}}\right)$
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  7. #7
    Junior Member Walagaster's Avatar
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    Re: finding a complex numbers

    Yes. I was wrong about the book being wrong.
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  8. #8
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    Re: finding a complex numbers

    Very neat! Thanks Idea.
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  9. #9
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    Re: finding a complex numbers

    Another way to see this is that $\cos{(x)} = \cos{(-z)}$. This holds in the complex domain too because the MacLaurin series contains only even powers of the variable.

    Thus, if $z=z_0$ is an answer, $z=-z_0$ must also be an answer.
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