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Natural logarithm of negative numbers

  1. Feb 17, 2005 #1
    I'm puzzled by....

    [tex]\ln \left( -\frac{1}{2} \right) = - \ln \left( 2 \right) + i \pi [/tex]

    Why is this true? How can I possibly get this result?

    I know that

    [tex]\ln \left( \frac{1}{2} \right) = - \ln \left( 2 \right)[/tex].

    Thank you so much
     
  2. jcsd
  3. Feb 17, 2005 #2

    dextercioby

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    Use Euler's formula and write "-1/2" in another way involving complex exponentials.Then apply "ln".

    Daniel.
     
  4. Feb 17, 2005 #3

    Hurkyl

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    Technically, that's not true. ln is a multivalued function. In particular,

    [tex]
    \ln -\frac{1}{2} = \ln \frac{1}{2} + i \pi (1 + 2n) \quad (n \in \mathbb{Z})
    [/tex]

    Here is the general formula, which can be derived as Daniel suggests.

    [tex]
    \ln z = \ln |z| + i \arg z
    [/tex]

    (where the ln on the RHS is the ordinary real-valued logarithm)
     
  5. Feb 17, 2005 #4

    dextercioby

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    Hurkyl,i was suggesting:

    [tex] -\frac{1}{2}=-1 \cdot \frac{1}{2} [/tex]

    And of course:

    [tex] -1=e^{i\pi (2n+1)} \ (n\in \mathbb{Z}) [/tex]

    No need to know any definition,just Euler's formula...:tongue2:

    Daniel.
     
  6. Feb 17, 2005 #5
    That makes sense! So, if

    [tex]z=-\frac{1}{2}=\frac{1}{2}\mbox{ cis }\pi (2n+1) = \frac{1}{2} e^{i\pi (2n+1)}[/tex]

    Then

    [tex]\ln z = \ln \left[ \frac{1}{2} e^{i\pi (2n+1)} \right] = -\ln 2\mbox{ } +\mbox{ } i\pi (2n+1) \qquad (n\in \mathbb{Z}) [/tex]

    Thanks
     
    Last edited: Feb 17, 2005
  7. Feb 18, 2005 #6

    dextercioby

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    Ha,that's interesting,u edited your message,yet u didn't see the typo "cis" instead of "cos"... :tongue:

    Daniel.
     
  8. Feb 18, 2005 #7

    Curious3141

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    FYI, [tex]cis \theta[/tex] is shorthand notation for [tex]\cos \theta + i \sin \theta[/tex]. It's perfectly valid.
     
  9. Feb 18, 2005 #8

    dextercioby

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    Hmmmmm...Interesting,even ingenious...

    Daniel.
     
  10. Feb 18, 2005 #9

    HallsofIvy

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    I haven't seen "cis" in years! (It's more engineer lingo than mathematics.)

    (And both dextercioby's posts were clearly tongue in cheek.)
     
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