Inconsistency in Complex LogarithmsDate: 06/08/2004 at 13:25:48 From: Eric Subject: Why ln(1) =/= ln(-1) + ln(-1)? I'm having difficulty with negative logarithms. I know ln(-1) = pi*i, and ln(1) = 0. But if I think of ln(1) as ln(-1 * -1) and use the property ln(ab) = ln(a) + ln(b), I get this: 0 = ln(1) = ln(-1 * -1) = ln(-1) + ln(-1) = pi*i + pi*i = 2pi*i =/= 0 What am I doing wrong? Date: 06/08/2004 at 14:15:52 From: Doctor Vogler Subject: Re: Why ln(1) =/= ln(-1) + ln(-1)? Hi Eric, Remember that in the real numbers, you can't take the log of a negative or zero, only of a positive. So you're talking about complex logarithms. Inverse functions sometimes behave in funny ways in the complex numbers. The square root does, too. For example, sqrt(-1) * sqrt(-1) =/= sqrt(1). In complex numbers, sqrt(-1) = i, and i*i = -1, but the square root of -1*-1 is 1. Well, you see, actually there are two square roots of 1. In fact, there are two square roots of any number (except zero, which has only one). If r is one square root, then -r is the other. In real numbers, we just always pick the positive root, but in complex numbers, it's not clear which root should be preferred. If we have to pick one, then we might choose to take the one in the first or fourth quadrant (with positive real part), or on the positive i axis in the case of pure imaginary roots. But, if we do that, then we lose the equality sqrt(a) * sqrt(b) = sqrt(ab). In fact, no matter how we make the choice, if we decide which square root to take as "the" square root for all complex numbers, then there will be two complex numbers with sqrt(a) * sqrt(b) = sqrt(ab), and two other complex numbers with sqrt(a) * sqrt(b) = -sqrt(ab). So we have a problem. Exactly the same thing happens with the complex logarithm. When you say ln(x), you mean the complex number y such that e^y or exp(y) = x. There is not just one but INFINITELY MANY such complex numbers y. If y is one of them, then all of the others are y + 2*n*pi*i for any integer n. So which one do we pick? This is where the texts talk about "branches" of the natural logarithm, which is just the mathematician's way of saying the particular choice he makes of which logarithm is "the" logarithm. The properties of the logarithm function are slightly different depending on which branch you use. But, most importantly, no matter what branch you use, there is a break somewhere. In particular, no matter what branch you use, or no matter where the break is, there will be two complex numbers a and b such that ln(a) + ln(b) and ln(ab) are two *different* logarithms of the same number ab (which means that they differ by some integer multiple of 2*pi). So in complex numbers, the rule becomes [ln(a) + ln(b) - ln(ab)]/[2*pi] is an integer. See also MathWorld's description at http://mathworld.wolfram.com/NaturalLogarithm.html Note that it speaks of the natural logarithm as a "multivalued function" {in fact, if you take all values of ln(a) and add them to all values of ln(b), then you will get all values of ln(ab)} and says it therefore requires a "branch cut" like I was talking about. The most natural place to put the cut is on the negative real axis (note the red line on their graph), and when you put the branch cut there, this particular logarithm is called the "principal value," which is the logarithm used in Mathematica (made by the owners of the MathWorld site). If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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