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Math Help - [SOLVED] The complex logarithm of the square root of i.

  1. #1
    Member Haven's Avatar
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    [SOLVED] The complex logarithm of the square root of i.

    Simple question but I can't seem to get it.

    Show that the set of values of log(i^{1/2}) is (n + 1/4)\pi\\i for any integer n.

    But using the definition of a complex logarithm and the polar form of i. I get:

    i = e^{i\pi/2}
    So, i^{1/2} = e^{i\pi/4}

    Since log(z) = ln(r) + i(\theta + 2n\pi)
    So, log(i^{1/2}) = ln(1) + i(\pi/4 + 2n\pi) = i(\pi/4 + 2n\pi)

    which is not the correct answer, is there something i'm missing?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Haven View Post
    Simple question but I can't seem to get it.

    Show that the set of values of log(i^{1/2}) is (n + 1/4)\pi\\i for any integer n.

    But using the definition of a complex logarithm and the polar form of i. I get:

    i = e^{i\pi/2}
    So, i^{1/2} = e^{i\pi/4}

    Since log(z) = ln(r) + i(\theta + 2n\pi)
    So, log(i^{1/2}) = ln(1) + i(\pi/4 + 2n\pi) = i(\pi/4 + 2n\pi)

    which is not the correct answer, is there something i'm missing?

    Note that i=e^{i\left(\frac{\pi}{2}+2n\pi\right)}.

    So i^{\frac{1}{2}}=e^{i\left(\frac{\pi}{4}+n\pi\right  )}.

    Therefore, \log\left(i^{\frac{1}{2}}\right)=\log e^{i\left(\frac{\pi}{4}+n\pi\right)}=i\left(\tfrac  {\pi}{4}+n\pi\right).

    Does this make sense?
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  3. #3
    Member Haven's Avatar
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    Dear lord, that makes a lot of sense. It's so simple now. Thanks alot
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