Is the equation $\log[A/(-B)]=\log[(-A)/B]$ valid? A friend sent me these lines:
$$\log[A / (-B)] = \log[(-A) / B]$$
$$\log(A) – \log(-B) = \log(-A) – \log(B)$$
$$\log(A) – [\log(-1) + \log(B)] = \log(-1) + \log(A) – \log(B)$$
$$\log(A) – \log(B) - \log(-1) = \log(-1) + \log(A) – \log(B)$$
Something is not right, but we are not sure what.
Edit
My friend actually wrote $\textrm{Log}$, not $\log$. Sorry.
 A: Rules for logarithm hold true only when each argument is positive. If $B>0$ then $-B&lt0$, so that only one of $\log(A/(-B))=\log A - \log (-B)$ and $\log((-A)/B)=\log (-A)-\log B$ can be correct.
A: The problem is that $\log$ is actually a function from $\mathbb{C}^*$ to $\mathbb{C}/2\pi i\mathbb{Z}$.
In $\mathbb{C}/2\pi i\mathbb{Z}$ it is actually true that $\log(-1) = \pi i+2\pi i \mathbb{Z} = -\pi i + 2\pi i \mathbb{Z} = -\log(-1)$.
A: A common convention is that Log is the principal branch of the logarithm, the one with imaginary part in the interval $(-\pi, \pi]\newcommand{Log}{\operatorname{Log}}$, while log can refer to any branch.
Now $\text{Im}(\Log (A) - \Log(B)) \in (-2\pi, 2\pi)$.  If it is in $(-2\pi, 0]$ you can add $\pi = \text{Im}(\Log(-1))$ to it and stay in the interval $(-\pi,\pi]$, while if it is in $(0,2\pi)$ you must subtract $\pi$.  So the correct answer is
$$\Log(A/(-B)) = \Log((-A)/B) = \cases{ \Log(A) - \Log(B) + \Log(-1) & if $\text{Im}(\Log(A)) \le \text{Im}( \Log(B))$\cr \Log(A) - \Log(B) - \Log(-1) & otherwise\cr}$$ 
A: Only for positive argument? As I understand it, there is a logarithmic identity even for complex numbers:
$$\ln( z_1 \cdot z_2 ) = \ln(z_1) + \ln( z_2 )$$
Now, in particular, is this right?:
$$\ln\left[(-1) \cdot z \right]  = \ln(-1) + \ln(z)$$
A: The short answer is the usual one: if you try to apply an identity without paying attention to when that identity is valid, you're likely to get all sorts of nonsensical results.
It's fair to speculate that the identities may still hold in situations where you haven't checked... but as soon as you derive something suspicious, you should suspect your speculation was wrong.
That said, the equation you wrote is actually true in a certain sense: the complex logarithm, $\log$, is not a function: it's a "multi-valued function", and the set of values the left hand side can take is equal to the set of values the right hand side can take.
In particular, $\log(-1)$ has the values $(\pi + 2 \pi n) \mathbf{i}$, where $n$ ranges over all integers. It's easy to see that $\log(-1)$ and $-\log(-1)$ have the same set of values.
However, multi-valued functions are one of those things you should avoid like the plague until you actually learn them properly: it's way too easy to make mistakes. It's more "common to use the principal branch" of the complex logarithm, $\def\Log{\mathop{\mathrm{Log}}}\Log$, which really is an ordinary function.
But when you do, you have to pay attention to the fact that $\Log$ does not satisfy the identity $\Log(xy) = \Log(x) + \Log(y)$. However, we do know the difference between the left and right hand sides is either $0$, $2 \pi \mathbf{i}$, or $-2 \pi \mathbf{i}$. It's generally pretty easy to figure out which, once you understand what $\Log$ is computing.
As an aside, there are other variations on the logarithm where $\log(-1) = 0$. For these variations, the equation you wrote holds. :)
A: Using the convention of Maple for the choice of principal value, we get
$$\begin{align}
&\log(1/(-1)) = \log(-1) = i\pi,
\\ &\log(1) -\log(-1)=0-(i\pi) = -i\pi.
\end{align}$$
so your second line
$\log(A)–\log(−B)=\log(−A)–\log(B)$ is wrong.
A: By the way, it would be advisable to used standard symbols for standard cases. In particular, I suggest that $\log$ be used for real-valued logarithms, since the logarithm function in subsets of $\mathbb{C}$ is much different.
