How is $\ln(-1) = i\pi$? [closed]
How do I derive:
$\ln(-1)=i\pi$ and
$\ln(-x)=\ln(x)+i\pi$ for $x>0$ and $x \in\mathbb R$
Thanks for any and all help!
Solution 1:
See here to get justification: $$e^{i\pi} = -1$$ So $\ln(-1) := i\pi$ is reasonable. Note that $\ln(-1) = i(2k+1)\pi$ for some $k\in\mathbb Z$ is just as reasonable. It all comes down to the branch of $\ln$ wich is chosen.
For the second part enforce $\ln(ab) = \ln(a) + \ln(b)$ where $a = -1$ and $b=x$.
Solution 2:
take the exponential to get $$e^{ln(-1)}=e^{\pi i}$$ $$-1=e^{\pi i}$$ $$e^{\pi i}+1=0$$ this is Euler's identity