# Log rule restrictions

I've been working with logarithms of negative numbers and I've realized that most of the common log rules don't hold for $x<0$. While this is stated in all the proofs I can find, I can't seem to work out exactly why that assumption is necessary, and where the proof fails without it. Does anyone know a good example of where and why a common log rule proof fails without this assumption?

Ex (power rule):

$$\ln(-1)=\ln(e^{i\pi})=i\pi$$

But

$$0=\ln(1)=\ln((-1)^2)\neq2\ln(-1)=2i\pi$$

So the power rule has failed in this case. Why?

The logarithm of a complex number, unlike real numbers, is not uniquely defined. This happens since $e^{z+2\pi i}=e^z$ for any complex number $z$. So the logarithm of $z$ is defined "up to $2\pi i$", and $\log z$ is sometimes defined as the set of all "options" for the logarithm.
Since this is the case, we consider the principal value of the logarithm to be $$\mathrm{Log}\left(z\right)=\ln\left|z\right|+i\mathrm{Arg}\left(z\right)$$ where $\mathrm{Arg}\left(z\right)\in\left(-\pi,\pi\right]$ (this is defined for $z\neq 0$). In other words, we took only one branch of the logarithm. So the usual logarithm rules may not apply to $\mathrm{Log}$.
However, it is true that $\log\left(z_1\cdot z_2\right)=\log z_1+\log z_2$ (thinking of the logarithms as sets), as well as most usual rules. But one must take extra care when dealing with the complex logarithm.
Well, because the $\ln$ of negative numbers is defined up to multiples of $2\pi i$. We could as well say that $\ln(-1) = 3\pi i$, because $e^{3\pi i} = -1$.