Ln(-1) = 0 ?!

  1. ln(-1) = 0 ???!

    supposed that we have

    [tex] \ln(-1) [/tex]

    then
    [tex] \frac{2}{2}\ln(-1) [/tex]

    so

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

    this is equal to

    [tex] \frac{1}{2}\ln(1) [/tex]

    and if this is equal to 0 the we can say that

    [tex] ln(-1) = 0 [/tex]

    is this right , wrong, are there any explanations for this?
     
  2. jcsd
  3. cristo

    cristo 8,394
    Staff Emeritus
    Science Advisor

    The rule of logarithms is a lnx=lnxa. In this case, a=1-- you cannot split it into a fraction and then only take the numerator!

    If you look at the logarithm graph, you will see that the function is not defined for negative x.
     
  4. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    It's wrong. ln(-1) is no longer a real number, so you can't treat it like one. This is like saying sqrt(-1) = (-1)1/2 = (-1)2/4 = ((-1)2)1/4 = 11/4 = 1.
     
  5. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    Actually, that step is perfectly valid in general - (a/a) ln(x) = 1/a ln(x^a), i.e. when everything is defined. Your next line explains why it's not valid here:
     
  6. cristo

    cristo 8,394
    Staff Emeritus
    Science Advisor

    Course it is; sorry!
     
  7. Really? Why can't one say that ln(-1)= i pi, for e^(i pi) = -1.
    Or is there something wrong with that line of logic?
     
  8. logarithm is defined also for complex numbers.
    ln(z)=ln(abs(z))+i*arg(z), where z is complex number, abs(z) is complex norm of complex number z, and arg(z) is its argument.
    So if -1 is treated as complex number -1+0*i, expression ln(-1) gives sense, but the identity a*ln(z)=ln(z^a) is no longer true.
     
  9. cristo

    cristo 8,394
    Staff Emeritus
    Science Advisor

    To satisfy the pedants, I shall re-phrase my above answer. The natural logarithm function, whose argument is a real number and to whom we can apply the standard laws of logarithms, is not defined for negative real numbers.
     
  10. for complex z: Ln(z) = ln(|z|) + i*Arg(z)

    so ln(-1) = ln(|-1|) + i*Arg(-1) = i*pi
     
    Last edited: Mar 6, 2007
  11. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    Really? If you were attempting to define the principal branch of Ln, then it ought to be Ln(z) = ln(|z|) + iArg(z), where ln is just the natural logarithm on the reals.

    In this case, we have Ln(-1) = ln(|-1|) + iArg(z) = i*pi.
     
  12. eh forgot the ln, fixed
     
  13. HallsofIvy

    HallsofIvy 40,218
    Staff Emeritus
    Science Advisor

    Cristo said, "for negative x". Since the complex numbers are not an ordered field, there are no "negative" complex numbers. Cristo was clearly talking about real numbers.
     
  14. Gib Z

    Gib Z 3,348
    Homework Helper

    [tex]\log_e -1 = i\pi + 2ki\pi, k\in \mathbb{Z}[/tex] Case Closed.
     
  15. is the logarithm of complex numbers defined for any other base except 'e'?
     
  16. Yes, you can still change between various different bases for your logarithms in the same manner as you do for Real numbers.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook
Similar discussions for: Ln(-1) = 0 ?!
Loading...