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Math Help - Solving |e^z| using complex numbers

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

    Solve |e^z| = 2 for z

    Attempt:

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

    Therefore:

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

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

    but the correct answer was z = ln2 + ik
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  2. #2
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    You need to notice that \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 \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:

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

    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:

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

    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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  5. #5
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    Oh I see, but why is it that you add iy to the solution and not 2 \pi k i?
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  6. #6
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    Quote Originally Posted by SyNtHeSiS View Post
    Oh I see, but why is it that you add iy to the solution and not 2 \pi k i?
    No, see post #2.
    The answer is z=\ln(2)+yi for any 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 z=\ln(2)+yi for any y\in \mathbb{R}.
    I understand what you did but can you just tell me why this method is wrong:

    2 = 2(cos0 + isin0)
    = 2e^0

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

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

    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 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 2\pi k i to \theta and in this case my \theta is 0, so then I did the following: = 2e^{0 + 2\pi k i}
    e^x = e^{ln2}e^{2\pi k i}
    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 2\pi ki here
    I don't know what your notes say, of course.
    But you are missing the whole idea here.

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