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Natural Log of negative number

  1. Dec 15, 2007 #1
    I've got a simple question that's been bugging me for a while. I think I know where the problem is, I'd just like a formal mathematical reason why I can't say this:

    [tex] \ln{(-1)}^2 = \ln(1) = 0 [/tex]

    That part is fine...but then:

    [tex] \ln{(-1)}^2 = 2 \ln(-1) = 2 (i \pi) [/tex]

    when they should obviously be equal.

    It presumably displays the fact that you can't take the square and put it in front of the log as a "2" when you're dealing with logs of negative numbers. I'd like to know if there's any formal theory behind why this can't be done.

  2. jcsd
  3. Dec 15, 2007 #2

    Ben Niehoff

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    What you're running into is the fact that [itex]\ln z[/itex] is multi-valued. Define [itex]\ln z[/itex] as the contour integral

    [tex]\ln z = \int_{\gamma} \frac{d\zeta}{\zeta}[/tex]

    where [itex]\gamma[/itex] is some contour running from 1 to z.

    Now, note that for each time the contour [itex]\gamma[/itex] winds around the origin in a positive sense, you get [itex]2\pi i[/itex] added to the integral (you can show this by the method of residues). Therefore, [itex]\ln z[/itex] is multivalued:

    [tex]\ln z = \ln_p z + 2n\pi i[/tex]

    where [itex]\ln_p z[/itex] is the principle value, and [itex]n[/itex] is any integer.
  4. Dec 15, 2007 #3


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    A simple way of seeing this is noting that exp(2n[pi]i)=1. Therefore ln(1)=2n[pi]i where n is any integer.
  5. Dec 15, 2007 #4
    Ah great! That makes perfect sense.

    So the natural log of 1 is, infact, multivalued if you allow a mapping to the complex plane?

    Thanks for cleaning that up anyway Ben & mathman.
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