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Math Help - natural log and imaginary unit question

  1. #1
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    natural log and imaginary unit question

    Hi,

    What happens when you take the natural log of a number that involves i?

    for eg.:

    \ln{ix}

    or even better:

    i\ln{ix}

    Or, put another way, what happens when you take the integral of a number involving i?

    \ln{ix}=\int_{1}^{x}\frac{1}{ix}dx

    Thanks
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  2. #2
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    Quote Originally Posted by rainer View Post
    What happens when you take the natural log of a number that involves i?
    First it is \log(z)=\log(x+yi).
    The definition of complex logarithm function is \log(z)=\ln(|z|)+i\cdot \arg(z).
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  3. #3
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    I think you misplaced a parenthesis but I got it anyway. Thanks.
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  4. #4
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    Quote Originally Posted by rainer View Post
    I think you misplaced a parenthesis but I got it anyway. Thanks.
    There is no missplaced parenthesis.
    \log(1-i)=\ln(\sqrt2)+i\left(\frac{-\pi}{4}+2k\pi\right)
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  5. #5
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    Ok, so if you didn't misplace a parenthesis I'm still confused.

    I'll just cut to the chase. I need help with this:

    \ln{i(\cosh1+\sqrt{\left|1-\cosh1\right|})}


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  6. #6
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    The principal value of \log\left(\cosh(1)+\sqrt{|1-\cosh(1)|}\right)= \ln\left(\cosh(1)+\sqrt{|1-\cosh(1)|}\right)+\frac{i\pi}{2}.
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