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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?

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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.

For more information, you may read here.

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  • $\begingroup$ Thanks! That was very informative. I was missing a whole layer of depth on the topic. $\endgroup$ – superckl May 2 '17 at 8:51
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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 $e^{3\pi i} = -1$.

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