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Math Help - Complex logarithm question

  1. #1
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    Complex logarithm question

    The question:
    Find the Cartesian form of Log((e^{\frac{3\pi i}{4}})^2) (note that this is the principle branch logarithm)

    My attempt:
    I'm not sure how to deal with the square. I recall that the index and log laws do not hold in this context. So how do I deal with it?
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:
    Find the Cartesian form of Log((e^{\frac{3\pi i}{4}})^2) (note that this is the principle branch logarithm)

    My attempt:
    I'm not sure how to deal with the square. I recall that the index and log laws do not hold in this context. So how do I deal with it?
    Isn't it true that \displaystyle e^{\left(\frac{3\pi i}{4}}\right)^2 = e^{\frac{3 \pi i}{2}} ....
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  3. #3
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    Ahh yes, now that I double checked, the index law does hold, but the log laws don't. I'm think I'm confusing principle value exponents with exponentials. :/

    I continued and did this:

    Log(e^{\frac{3\pi i}{2}})
    = ln|1| + (\frac{3}{2} + 2k)\pi i : k \in Z
    However it's the principle log, so k = 0.

    = \frac{3\pi}{2}i

    However, the solution is \frac{-i\pi}{2}

    Any suggestions?
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  4. #4
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    Quote Originally Posted by Glitch View Post
    Ahh yes, now that I double checked, the index law does hold, but the log laws don't. I'm think I'm confusing principle value exponents with exponentials. :/

    I continued and did this:

    Log(e^{\frac{3\pi i}{2}})
    = ln|1| + (\frac{3}{2} + 2k)\pi i : k \in Z
    However it's the principle log, so k = 0.

    = \frac{3\pi}{2}i

    However, the solution is \frac{-i\pi}{2}

    Any suggestions?
    Well, the principle log would require using the principle argument, would it not? Look up what definition of principle argument you're meant to be using.
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  5. #5
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    I think I understand now. Thank you.
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