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Thread: find all values of z satisfying...

  1. #1
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    find all values of z satisfying...

    $\displaystyle exp[2z] + exp[z] + 1 = 0 $

    Really not sure how to start this one, i tried breaking it into it's real and imaginary components but didn't get far with that.

    $\displaystyle exp[2x] \cdot(cos(2y)+isin(2y)) + exp[x]\cdot(cos(y)+isin(y)) +1 = 0 ?? $

    Is it something to do with taking the log?
    If so i'm a little lost on how to do it
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    Quote Originally Posted by linalg123 View Post
    $\displaystyle exp[2z] + exp[z] + 1 = 0 $

    Really not sure how to start this one, i tried breaking it into it's real and imaginary components but didn't get far with that.

    $\displaystyle exp[2x] \cdot(cos(2y)+isin(2y)) + exp[x]\cdot(cos(y)+isin(y)) +1 = 0 ?? $

    Is it something to do with taking the log?
    If so i'm a little lost on how to do it
    Start by substituting w = e^z and solving for w.
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    $\displaystyle w^2 + w + 1 = 0 $

    $\displaystyle w = \frac{-1 \pm \sqrt{-3}}{2}$

    $\displaystyle w = \frac{-1 \pm 3i}{2}$

    $\displaystyle z = \ln{\frac{-1 \pm 3i}{2}}$

    not sure how to continue?
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  4. #4
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    Quote Originally Posted by linalg123 View Post
    $\displaystyle w^2 + w + 1 = 0 $

    $\displaystyle w = \frac{-1 \pm \sqrt{-3}}{2}$

    $\displaystyle w = \frac{-1 \pm 3i}{2}$

    $\displaystyle z = \ln{\frac{-1 \pm 3i}{2}}$

    not sure how to continue?
    Write the numbers in polar form

    $\displaystyle \frac{-1+i\sqrt{3}}{2}=e^{i\frac{5\pi}{6}} \implies e^{z}=e^{i\frac{5\pi}{6}} $
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    Quote Originally Posted by linalg123 View Post
    $\displaystyle w^2 + w + 1 = 0 $

    $\displaystyle w = \frac{-1 \pm \sqrt{-3}}{2}$

    $\displaystyle w = \frac{-1 \pm 3i}{2}$

    $\displaystyle z = \ln{\frac{-1 \pm 3i}{2}}$

    not sure how to continue?
    Surely you've been taught about the complex log function ....?

    You should know that $\displaystyle \ln (z) = \ln |z| + i (\theta + 2n \pi)$ where $\displaystyle \theta$ is the principle argument of z and n is an integer. (The z here is not to be confused with the z in your question, by the way). Go back and review your class notes and textbook.
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    Quote Originally Posted by linalg123 View Post
    $\displaystyle w^2 + w + 1 = 0 $

    $\displaystyle w = \frac{-1 \pm \sqrt{-3}}{2}$

    $\displaystyle w = \frac{-1 \pm 3i}{2}$

    $\displaystyle z = \ln{\frac{-1 \pm 3i}{2}}$

    not sure how to continue?
    There is no need to use logarithms, if you note that $\displaystyle \displaystyle w = e^{z}$ and that you can write each of the values of $\displaystyle \displaystyle w$ in exponential form.
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    when i try that i get a messy answer.

    $\displaystyle |z|= \sqrt{(\frac{-1}{2})^2 + (\frac{3}{2})^2)} = \sqrt{\frac{10}{4}}$

    $\displaystyle \theta = \arctan{\frac{\frac{3}{2}}{\frac{1}{2}}} = \arctan{3}$

    did i do something wrong?
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  8. #8
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    Yes, the imaginary part of $\displaystyle \displaystyle w$ is $\displaystyle \displaystyle \pm \frac{\sqrt{3}}{2}$, not $\displaystyle \displaystyle \pm \frac{3}{2}$.
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    well that was dumb, thanks.

    so it's $\displaystyle z= i(2n\pi \pm \frac{2\pi}{3}) $ ?
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