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Thread: Natural log of a negative number

  1. #1
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    Natural log of a negative number

    I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there.

    ln(-.3675)
    ln(-13.4)

    Here are a few problems from my book, can anyone guide me through?
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  2. #2
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    Quote Originally Posted by Failbait View Post
    I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there.

    ln(-.3675)
    ln(-13.4)

    Here are a few problems from my book, can anyone guide me through?
    If $\displaystyle z\not = 0$ then $\displaystyle \log z = \log |z| + i\arg (z)$

    Can you do that?
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  3. #3
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    ln(-.3675) = $\displaystyle -1.001+\pi i$
    ln(-13.4) = $\displaystyle 2.595+\pi i$

    Do those answers look about right?

    Thanks!
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  4. #4
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    Quote Originally Posted by Failbait View Post
    ln(-.3675) = $\displaystyle -1.001+\pi i$
    ln(-13.4) = $\displaystyle 2.595+\pi i$

    Do those answers look about right?

    Thanks!
    Hmmmm, let's see now ..... does |-0.3675| = -1.001? Does |-13.4| = 2.595?
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  5. #5
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    $\displaystyle e^{i\theta} = cos \theta + i sin \theta$
    $\displaystyle e^{i\pi} = cos \pi + i sin \pi$
    $\displaystyle e^{i\pi} = -1$
    $\displaystyle ln(e^{i\pi}) = ln -1$
    $\displaystyle i\pi = ln -1$

    For $\displaystyle x \in R^+$
    $\displaystyle ln -x = ln -1 + ln x = ln x + i\pi$

    This is the same thing as ThePerfectHacker's formula. Remember that mine only works for real numbers.

    ln(-.3675) = $\displaystyle -1.001+\pi i$
    ln(-13.4) = $\displaystyle 2.595+\pi i$

    Yes, they're correct.
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  6. #6
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    Quote Originally Posted by wingless View Post
    $\displaystyle ln(e^{i\pi}) = ln -1$
    $\displaystyle i\pi = ln -1$
    You are not allowed to do that. The rule $\displaystyle \log e^z = z$ for $\displaystyle z\in \mathbb{C}^{\text{x}}$ does not work in general.
    However, $\displaystyle e^{\log z} = z$ does work in general.
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  7. #7
    Super Member wingless's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You are not allowed to do that. The rule $\displaystyle \log e^z = z$ for $\displaystyle z\in \mathbb{C}^{\text{x}}$ does not work in general.
    Wow, I didn't know that. Sorry for that mistake, thanks for correcting me. Ok, the formula is still true since $\displaystyle ln -1 = i \pi$.
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  8. #8
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    Quote Originally Posted by wingless View Post
    Ok, the formula is still true since $\displaystyle ln -1 = i \pi$.
    Yes! Sometimes the formula will work sometimes it does not.
    Another thing that breaks down is $\displaystyle \log (z_1z_2) = \log z_1 + \log z_2$.
    Take for example $\displaystyle z_1=z_2=-1$.
    Complex logarithm.
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