derlingasmh

2022-01-25

Natural log of a negative number
$\mathrm{ln}\left(-1\right)=\mathrm{ln}\left({e}^{i\pi }\right)=i\pi$

Tapanuiwp

Expert

Context is important here. In the context of real numbers, negative numbers have no logarithms (and neither does 0) because $\mathrm{log}\left(x\right)$ is a number y such that ${e}^{y}=x$ and ${e}^{y}$ is always greater than $0$.
On the other hand, in the context of complex numbers, every complex number other than 0 has logarithms. In fact, any such complex number has infinitely many logarithms! You are right when you claim that $i\pi$ is a logarithm of $-1$. However, every complex number of the form $\pi i+2\pi \in$ (with $n\in \mathbb{Z}$) is also a logarithm of $-1$, since
${e}^{\pi i+2n\pi i}={e}^{\pi i}{e}^{2\pi \in }=\left(-1\right)×1=-1.$

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