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Is this formula justified?

  1. Jun 23, 2012 #1
    I've read about complex logarithms and maybe it's just my reading skill, but it all seems to be way too complicated. Anyways, after doing some thinking of my own I've come up with a few formulas, one of which is ln(-a) = ln(a) + πi; however, my reasoning behind such formulas seems way too basic, so I'm doubting myself. Could someone please check my work and let me know if what I'm doing is ok?

    Here's the derivation:
    ln(-a) = ln(-a)
    ln(-a) = ln(-1a)
    ln(-a) = ln(-1) + ln(a) -- Properties of Logs
    ln(-a) = πi + ln(a) -- Euler's Identity

    Euler's identity states e^(iπ) = -1; therefore, ln(-1) = iπ.
     
  2. jcsd
  3. Jun 23, 2012 #2

    Char. Limit

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    So long as you restrict a to being positive and real, that formula should be valid. Well, up to a multiple of 2πi anyway.
     
  4. Jun 23, 2012 #3
    Wait a minute, that's an excellent point.. Why does the output of the function change depending on how you organized the terms?

    ln(-a) = πi + ln(a)
    If we attempt to let a = -1
    ln(1) = πi + ln(-1)
    ln(1) = 2πi

    However, if we write ln(-1) = πi + ln(a) as
    ln(a) = ln(-a) - πi
    Now if we let a = 1 to solve for ln(1)
    ln(1) = ln(-1) - πi
    ln(1) = πi - πi
    ln(1) = 0

    Is 2πi the same as zero? Looking at Euler's Identity I've never seen the π as evidence of a cycle, I've always just thought of it as a constant. Interesting... That's something to think about..

    e-
    The formula is intended to work for all a, a != 0. Why must a be limited to positive values?
     
  5. Jun 23, 2012 #4

    micromass

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    The complex logarithm is a tricky animal. First of all, it is multivalued. This can easily be seen by

    [tex]e^0=1=e^{2\pi i}[/tex]

    So the logarithm of 1 should be both 0 and [itex]2\pi i[/itex] (and other values).

    A formula for the complex logarithms can be given by

    [tex]Log(z)=log|z| + i arg(z) + 2\pi i k[/tex]

    for all [itex]k\in \mathbb{Z}[/itex] (that is: each k gives a good value for the logarithm).

    The arg(z) is the angle that z makes with the positive x-axis in radians.

    For example, if z is a positive real, then it makes a 0 angle with the positive x axis, and thus

    [tex]Log(z)=log(z)+2\pi i k[/tex]

    if k=0, then this corresponds with the ordinary logarithm.

    If z is a negative real, then it makes a [itex]\pi[/itex] (or [itex]-\pi[/itex], doesn't matter) angle with the positive x-axis. And thus

    [tex]Log(z)=log(-z)+\pi i + 2\pi i k[/tex]

    If k=0, then we get the formula you discovered.

    Mathematicians don't like multi-valued functions, that's why they restrict the logarithm to only take on one value. That is: we always take k=0 in the previous formula. This is called the principal branch of the logarithm. This is a very arbitrary choice and other choice can be made. But it has as benefit that the complex logarithm coincides with the real logarithm for positive real numbers.

    Also, complex logarithms are dangerous because many familiar laws do not hold anymore. For example, Log(zw)=Log(z)+Log(w) is not valid anymore (certainly not for the principal branch logarithm).
     
  6. Jun 23, 2012 #5
    Hmm, so I should be sure when using properties of logs that I don't perform operations on imaginary logs unless I can justify the formula in the imaginary plane first?

    I've seen this formula before but it's one of those ones where I take a look at it and shy away, but really it's not all that difficult. This post will surely give me a lot to think about. Thank you

    (I think what I'm going to do next is try to generalize the familiar log formulas.)

    P.s. The log above is considered to be the natural log right? Or is that formula a generalization for all bases?

    e-
    Two questions:
    Is this operation ok?
    ln(a) = ln(-1) + ln(-a)?
    This is a basic rule of logarithms, but does the ln(-1) make the use of the rule questionable?

    Do you have a proof of that formula? I'd like to see how it was derived.
     
    Last edited: Jun 23, 2012
  7. Jun 23, 2012 #6

    micromass

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    Yes. I think most properties will hold for the multivalued log, but not for the principal branch. But it needs to be proven first.

    It's really not hard. Take a complex number w. Denote

    [tex]Log(w)=z[/tex]

    this means by definition that

    [tex]e^z=w[/tex]

    Write [itex]z=z_1+iz_2[/itex]. We can write w in its polar form as [itex]|w|(\cos\theta +i\sin\theta)[/itex] where [itex]\theta[/itex] is the arg of w. Thus we have

    [tex]e^{z_1+iz_2}=|w|(\cos\theta+i\sin\theta)[/tex]

    By Eulers formula, we have

    [tex]e^{z_1+iz_2}=e^{z_1}e^{iz_2}=e^{z_1}(\cos z_2 + i\sin z_2)[/tex]

    So we got the equation

    [tex]e^{z_1}(\cos z_2 + i\sin z_2)=|w|(\cos\theta +i\sin\theta)[/tex]

    We deduce from this that [itex]e^{z_1}=|w|[/itex] and thus [itex]z_1=log|w|[/itex]. Also, we have that

    [tex]\cos z_2 = \cos\theta~\text{and}~\sin z_2=\sin\theta[/tex]

    This implies that [itex]z_2=\theta+k2\pi[/itex] for a certain k. Thus

    [tex]Log(w)=log|w|+i(\theta+k2\pi)[/tex]

    which is the formula we wanted.

    That is a formula for the natural log. I'm sure there are generalizations for other bases. But for some reason, other bases don't really show up much in complex analysis.
     
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