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Math Help - 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 log((1 + i)(1 - i))?
    (and I'm going to assume this is log_e)

    = log(1 + i) + log(1 - i)

    First:
    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
    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:
    log(1 - i) = \frac{log(2)}{2} - i\frac{\pi}{4}

    So
    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:
    log \left [(1 + i)^{1 - i} \right ]

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

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

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

    -Dan
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  5. #5
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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 \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 \log (xy) = \log x + \log y \mbox{ and }\log x^y = y\log x does not work in \mathbb{C} for x,y\not = 0 in general.
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