# Inequality, what is wrong with $\log(-1) = - \log(-1)$?

Can anyone tell me what is wrong with the following line of argument:

$$\log(-1) = \log(-1) - \log(1) = - \bigg( \log(1) - \log(-1) \bigg) = - \log \Big( \frac{1}{-1} \Big) = - \log(-1)$$

Considering the complex logarithm the left-hand-side evaluates to $i \pi$ and the right-hand-side evaluates to $- i \pi$.

• Logarithm is multi-valued on the complex numbers, see wiki. – vadim123 Oct 8 '13 at 14:23
• Any branch of the logarithm on $\mathbb C$ does not have the property that $\log ab = \log a + \log b$ for all $a$, $b$ – martini Oct 8 '13 at 14:24
• What is wrong with $1 = \sqrt{1} = \sqrt{\frac{-1}{-1}} = \sqrt{\frac{-1}{1}} \sqrt{\frac{1}{-1}} = \sqrt{{-1}} \sqrt{{-1}} = -1$? Treating complex numbers as if they were real and positive. – Henry Oct 8 '13 at 14:32
• @vadim123 But he uses $\log$ as a function so this is an argument suggesting that the equality is not well-formed formula rather than it doesn't hold. – user87690 Oct 8 '13 at 14:35
• I would assume that any answer to a question like this will start with a claim of which equality is actually incorrect. :-) – user87690 Oct 8 '13 at 14:42

It's because the logarithm, if you extend it to the complex numbers, is only properly defined on a Riemann surface.

What does that mean? First of all, notice that you're getting $i\pi$ on the left hand side, and $-i\pi$ on the right hand side. It is interesting to notice that:

$$e^{i\pi}=-1=e^{-i\pi}$$

Now the logarithm can be naively defined as the 'inverse of the exponential'. So why do we say $\log(-1)=i\pi$ and not $\log(-1)=-i\pi$?

Well, first notice that $e^{2i\pi}=1$. So $e^a=e^{a+2ni\pi}$ for any integer $n$. This means that the logarithm is only defined up to addition of $2ni\pi$. There are two ways round this:

1. Just choose a value for the logarithm at each point. Remember that the logarithm of a complex number $re^{i\theta}$ is given by $\log_e(r)+i\theta$, where $\theta$ is the argument. It is common to specify that, for example, $-\pi<\theta\le\pi$. This is called a branch cut, because we cut along the negative real axis, and end up with one branch of the complex logarithm - another branch is given by $\pi<\theta\le3\pi$. For the first branch $\theta$ for $-1$ is $\pi$, so we say that $\textrm{Log}(\theta)=i\pi$ (the capital L on $\textrm{Log}$ refers to the principal value of the logarithm - what we get if we restrict ourselves to a particular branch. Notice that this logarithm coincides with the real logarithm if and only if the argument for real numbers is chosen to be $0$ and not $2n\pi$ for some integer $n\ne 0$.

2. The branch cut means that the logarithm is not continuous on the negative real axis (indeed, it is usually not defined along the cut at all). Instead of studying the complex logarithm over the complex numbers, we introduce a Riemann surface - a one-dimensional complex manifold that 'joins up' the branch cuts so that the logarithm is now continuous everywhere. This picture shows you sort of what $\log$ looks like on a Riemann surface: Unfortunately, in neither of those cases can you assume that usual relationships like $\log(ab)=\log(a)+\log(b)$ still hold. For example, if we make a branch cut along the negative real axis, we could try to work out $\log((-1+i)\times(-1+i))=\log(-2i)$ by writing it as $2\log(-1+i)$. The argument $\theta$ for $-1+i$ is $3\pi/2$, but doubling that angle 'crosses' the branch cut, and we end up with $-2i$, which has an argument not of $3\pi$ but of $-\pi$. Of course, we could place the cut somewhere else and we'd be fine. In your example, though, you will end up crossing the branch cut wherever you put it.

Why is that? Your example is basically equivalent to saying that:

$$\log(-1)=\log\left(\frac{1}{-1}\right)=-\log(-1)$$

Now $-1$ can be thought of as going round the unit circle by an angle of $\pi$. The reciprocal of a complex number always has negative of the argument of the original number, so $\dfrac{1}{-1}$ can be thought of as going round the unit circle by an angle of $-\pi$; i.e., you reach the same point travelling in the opposite direction. Between them, the two paths cover the entire unit circle. Since you've got to place a branch cut somewhere on the circle, you're going to end up crossing it at some point, so you've got to get an answer that makes no sense.

Update: I can't remember where I first read this, but it's definitely true. It's easy to look at something like this and conclude that it's a slightly annoying area of mathematics that you have to deal with even though you'd rather it worked a different way. In fact, the opposite is true: this seemingly annoying property of the complex logarithm in fact opens up many beautiful areas of mathematics, including the study of Riemann surfaces.

• An aside: when you get to the point where you can keep track of which branch everything is on,then the answer does make sense. Relationships between different branches can be useful to know. – user14972 Oct 8 '13 at 23:33

All the equalities but one are trivial. But how do you know that $\log(1) - \log(-1) = \log(\frac{1}{-1})$? It actually doesn't hold. As @martini says it doesn't hold in general that $\log(ab) = \log(a) + \log(b)$.

When you take the Logarithms of Negative numbers, you need to be very careful, some of the common rules of logarithms do not apply with Negative Logarithms Let's assume we integrate "from $-1$ to $1$" and along a semicircle ( of radius $\epsilon\,, \quad 0 < \epsilon < 1$ ) below the $x$-axis: $$\int_{C}{{\rm d}z \over z} = \int_{-1}^{-\epsilon}{{\rm d} x \over x} + \int_{-\pi}^{0} {\epsilon{\rm e}^{{\rm i}\theta}{\rm i}\,{\rm d}\theta \over \epsilon{\rm e}^{{\rm i}\theta} } + \int^{1}_{\epsilon}{{\rm d} x \over x} = {\rm i}\pi$$ and $${\rm i}\pi = \lim_{\epsilon \to 0^{+}}\int_{C}{{\rm d}z \over z}\ \mbox{and never}\ = \overbrace{\int_{-1}^{1}{{\rm d} x \over x}} ^{\mbox{It doesn't have any sense !!!.}}$$ The example emphasizes that integration in the Complex Plane requires the election of a path and it's not interpreted in any other way like $\displaystyle{\int_{-1}^{1}\ldots}$.

Your error is thinking that $\log(-1)$ evaluates to $\mathbf{i} \pi$!

$\log$ is a multi-valued function. The set of values for $\log(-1)$ is $(1 + 2n) \mathbf{i} \pi$, where $n$ ranges over all integers.

It's not hard to see that the negation of any value in this set is also in this set, so the set of values for $-\log(-1)$ is the same.

At the lowest level, the problem could be said to be that you weren't paying attention to the "branches" of the logarithm, and you mistakenly thought the two sides referred to the same branch. Or, it could be said to be that you were being inattentive of the details of the algebra you tried to do, and attempted to apply a logarithm identity in a situation where it doesn't work.

At a high level, though, the explanation is simple: you haven't really learned about multi-valued functions, and you've rediscovered one of the many problems that arise when you try to work with them in a naive fashion.

• Incidentally, "naive" is not meant to be an insult: it more or less simply means doing things in a straightforward manner based on how they appear (where that appearance is colored by prior experiences). And this works (to varying extents) for a lot of things. – user14972 Oct 8 '13 at 22:59

I think $\log(1)=0$ doesn't hold in the complex,so "$\log(-1)=log(-1)-log(1)$"is wrong.

• $\log(1) = 0$ holds even for complex function. Complex logarithm is defined as real logarithm of absolute value times the argument of the complex number. – user87690 Oct 8 '13 at 14:38
• @user87690 - It's defined as the real logarithm of the absolute value + $i$ times the argument. – John Gowers Oct 8 '13 at 14:41
• @Donkey_2009: Yes of course, sorry, I messed it up. But I thought it kind of right. :-) – user87690 Oct 8 '13 at 14:44
• Yes,I'm sorry.it is right,after all both the left and the right are multipvalued. – F.G. Oct 8 '13 at 15:03
• @user: $\text{Log}(1) = 0$: but $\log(1)$ is multivalued, and comprises all numbers of the form $2 \pi \mathbf{i} n$, where $n$ is an integer. – user14972 Oct 8 '13 at 23:30