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Ln (-1)

  1. Jan 18, 2012 #1
    I saw a post discussing (-1)(1/3). After I saw it, I was a little confused about how the value of ln(-1) is derived. I could see from euler's identity that it should equal i*π; however, I'm not sure where the multiplicity (2k+1) comes from. I searched the forums (and google) but couldn't find anything. Could anyone clear this up for me?
     
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  3. Jan 18, 2012 #2

    micromass

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    Well, saying that [itex]ln(-1)=(2k+1)i\pi[/itex] for [itex]k\in \mathbb{Z}[/itex] simply means (by definition) that

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

    Why does this hold? Well, by definition of the complex exponential, we have

    [tex]e^{(2k+1)i\pi}=\cos((2k+1)\pi)+i\sin((2k+1)\pi)[/tex]

    From elementary trigonometry, we know that [itex]\cos((2k+1)\pi)=-1[/itex] and [itex]\sin((2k+1)\pi)=0[/itex]. So we indeed see that the above expression equals -1.

    However: most authors don't like the logarithm to take on multiple values. They want one value for the complex logarithm. That's why you will always see [itex]ln(-1)=i\pi[/itex]. We don't mention the other values because we only want one answer for ln(-1). We say that [itex]i\pi[/itex] is the principal value of the logarithm.
     
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