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Thread: imaginary part

  1. #1
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    imaginary part

    Separate real and imaginary part of log (1+i)1-i.
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  2. #2
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    Quote Originally Posted by whackywheelz View Post
    Separate real and imaginary part of log (1+i)1-i.
    Is this $\displaystyle log((1 + i)(1 - i))$?
    (and I'm going to assume this is $\displaystyle log_e$)

    $\displaystyle = log(1 + i) + log(1 - i)$

    First:
    $\displaystyle 1 + i = \sqrt{2} \left [ cos \left ( \frac{pi}{4} \right ) + i~sin \left ( \frac{\pi}{4} \right ) \right ] = \sqrt{2}e^{i \pi/4} $

    So
    $\displaystyle log(1 + i) = log \left ( \sqrt{2}e^{i \pi/4} \right ) = log(\sqrt{2}) + log \left ( e^{i \pi/4} \right ) = \frac{log(2)}{2} + i\frac{\pi}{4}$

    Similarly:
    $\displaystyle log(1 - i) = \frac{log(2)}{2} - i\frac{\pi}{4}$

    So
    $\displaystyle log((1 + i)(1 - i)) = \frac{log(2)}{2} + \frac{log(2)}{2} = log(2)$

    (Of course the short way is to note that (1 + i)(1 - i) = 2, but that would spoil all the fun, wouldn't it? )

    -Dan
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  3. #3
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    correction

    Separate real and imaginary part of log (1+i)^1-i.

    i.e raised to (1-i). sorry bout that. the compose post view shows me correctly when i opy paste from word.


    thanks Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by whackywheelz View Post
    Separate real and imaginary part of log (1+i)^1-i.

    i.e raised to (1-i). sorry bout that. the compose post view shows me correctly when i opy paste from word.


    thanks Dan
    Ah, that makes a lot more sense now.

    The concept is pretty much the same problem as I worked out before:
    $\displaystyle log \left [(1 + i)^{1 - i} \right ] $

    $\displaystyle = (1 - i) log(1 + i) = (1 - i) \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

    $\displaystyle = \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right ) - i \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

    So the imaginary part is
    $\displaystyle - \left ( \frac{log(2)}{2} + \frac{\pi}{4} \right )$

    -Dan
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    thanks a ton!

    just one more doubt about the same question.

    which is the real part?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by whackywheelz View Post
    just one more doubt about the same question.

    which is the real part?
    The part that doesn't have an "i" in it, the $\displaystyle \frac{log(2)}{2} + \frac{\pi}{4}$. In this case the real and imaginary parts happen to be negatives of one another. This is not typical.

    -Dan
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  7. #7
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    Be careful the rule $\displaystyle \log (xy) = \log x + \log y \mbox{ and }\log x^y = y\log x$ does not work in $\displaystyle \mathbb{C}$ for $\displaystyle x,y\not = 0$ in general.
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