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Why does the graph of y = ln(x) have an imaginary part?

  1. Nov 29, 2011 #1
    I was messing around with wolframalpha and tried to make this graph.


    Now I understand why the blue real part exists and has that shape but I don't understand why it has an imaginary part?

  2. jcsd
  3. Nov 29, 2011 #2

    D H

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    In terms of the real numbers, the natural logarithm function maps positive reals to the reals. Given some negative real number x, there is no real number y such that ey=x (i.e., such that y=log(x).) The natural logarithm can be extended to the complex numbers via Euler's equation ex=cos(x)+i*sin(x). This leads to log(-1)=pi*i (And to a host of other values; natural log is no longer unique.)
  4. Nov 29, 2011 #3
    Ok, but where does imaginary pi come into it?

    I understand that from a bit of messing around that any value to the left of the y axis is found by taking the natural log and adding imaginary pi, but I don't understand where imaginary pi comes from.
  5. Nov 29, 2011 #4

    D H

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    It comes from Euler's identity: [itex]e^{\pi i}+1=0[/itex].
  6. Nov 29, 2011 #5


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    I don't know what math background you have, but the key result here is
    [itex]e^{ix} = \cos x + i \sin x[/itex]
    If you aren't familar with that, Google for "Euler's identity"

    Suppose [itex]z = \log(-a)[/itex] where [itex]a > 0[/itex]. Then [itex]e^z = -a[/itex]. Since [itex]e^x > 0[/itex] for any real value of [itex]x[/itex], [itex]z[/itex] must be complex so let [itex] z = x + iy[/itex].

    Then [itex]e^{x + iy} = e^x(\cos y + i \sin y) = -a[/itex]

    The imaginary part = 0, so [itex]\sin y = 0[/itex].

    So [itex]\cos y = \pm 1[/itex] and to make the real parts equal we have to take [itex]\cos y = -1[/itex] and [itex]e^x = a[/itex] or [itex]x = \log a[/itex].

    Putting it all together

    [itex]\log(-a) = \log(a) + i \pi + 2k\pi[/itex] where [itex]k[/itex] is any integer.

    The graph shows the "principal value" i.e. [itex]\log(a) + i \pi [/itex].
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