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Math Help - find all values of z satisfying...

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

     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.

     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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  2. #2
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    Quote Originally Posted by linalg123 View Post
     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.

     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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  3. #3
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     w^2 + w + 1 = 0

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

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

     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
     w^2 + w + 1 = 0

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

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

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

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

    \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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  5. #5
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    Quote Originally Posted by linalg123 View Post
     w^2 + w + 1 = 0

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

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

     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 \ln (z) = \ln |z| + i (\theta + 2n \pi) where \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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  6. #6
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    Quote Originally Posted by linalg123 View Post
     w^2 + w + 1 = 0

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

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

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

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

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

     \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 w is \displaystyle \pm \frac{\sqrt{3}}{2}, not \displaystyle \pm \frac{3}{2}.
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  9. #9
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    well that was dumb, thanks.

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