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Inconsistency in Complex Logarithms




Date: 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/ 
Associated Topics:
High School Imaginary/Complex Numbers
High School Logs

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