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Thread: Solving |e^z| using complex numbers

  1. #1
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    Solving |e^z| using complex numbers

    Solve $\displaystyle |e^z| = 2$ for z

    Attempt:

    $\displaystyle \sqrt{e^{2z}} = e^z$

    Therefore:

    $\displaystyle e^z = 2e^{0 + i2\pi k}$
    $\displaystyle = e^{ln2 + i2\pi k}$

    $\displaystyle z = ln2 +i2\pi k, k E Z$

    but the correct answer was $\displaystyle z = ln2 + ik$
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  2. #2
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    You need to notice that $\displaystyle \left| {e^z } \right| = \left| {e^{x + yi} } \right| = e^x \left| {\cos (y) + i\sin (y)} \right| = e^x $
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  3. #3
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    Quote Originally Posted by Plato View Post
    You need to notice that $\displaystyle \left| {e^z } \right| = \left| {e^{x + yi} } \right| = e^x \left| {\cos (y) + i\sin (y)} \right| = e^x $
    But then how can you now solve for z like the orignal question asked to, since you dont have a z variable anymore?

    Using what you told me I got:

    $\displaystyle e^x$
    $\displaystyle = 2$
    $\displaystyle = e^{ln2}e^{0 + 2\pi k i}$

    $\displaystyle x = ln2 + 2\pi k i, k E Z$

    but my answer obviously isnt right.
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    But then how can you now solve for z like the orignal question asked to, since you dont have a z variable anymore?

    Using what you told me I got:

    $\displaystyle e^x$
    $\displaystyle = 2$
    $\displaystyle = e^{ln2}e^{0 + 2\pi k i}$

    $\displaystyle x = ln2 + 2\pi k i, k E Z$

    but my answer obviously isnt right.
    You're expected to realise that x is real ....
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    Oh I see, but why is it that you add $\displaystyle iy$ to the solution and not $\displaystyle 2 \pi k i$?
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    Quote Originally Posted by SyNtHeSiS View Post
    Oh I see, but why is it that you add $\displaystyle iy$ to the solution and not $\displaystyle 2 \pi k i$?
    No, see post #2.
    The answer is $\displaystyle z=\ln(2)+yi$ for any $\displaystyle y\in \mathbb{R}.$
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  7. #7
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    Quote Originally Posted by Plato View Post
    No, see post #2.
    The answer is $\displaystyle z=\ln(2)+yi$ for any $\displaystyle y\in \mathbb{R}.$
    I understand what you did but can you just tell me why this method is wrong:

    $\displaystyle 2 = 2(cos0 + isin0)$
    $\displaystyle = 2e^0$

    I read somewhere in my notes that you always add $\displaystyle 2\pi k i $to $\displaystyle \theta$ and in this case my $\displaystyle \theta$ is 0, so then I did the following:
    $\displaystyle
    = 2e^{0 + 2\pi k i}$

    $\displaystyle e^x = e^{ln2}e^{2\pi k i}$

    $\displaystyle x = ln2 + 2\pi k i, k E R$

    I can see that k must be R since x is a real number.

    But I just want to know why you dont add $\displaystyle 2\pi ki$ here
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  8. #8
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    Quote Originally Posted by SyNtHeSiS View Post
    I read somewhere in my notes that you always add $\displaystyle 2\pi k i $to $\displaystyle \theta$ and in this case my $\displaystyle \theta$ is 0, so then I did the following: $\displaystyle = 2e^{0 + 2\pi k i}$
    $\displaystyle e^x = e^{ln2}e^{2\pi k i}$
    $\displaystyle x = ln2 + 2\pi k i, k E R$
    I can see that k must be R since x is a real number.
    But I just want to know why you dont add $\displaystyle 2\pi ki$ here
    I don't know what your notes say, of course.
    But you are missing the whole idea here.

    $\displaystyle \left|e^{\ln(2)+yi}\right|=\left|e^{\ln(2)}\right| \left|\cos(y)+i\sin(y)\right|=2$ for all $\displaystyle y\in \mathbb{R}$.
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