How is $\ln(-1) = i\pi$?

Question

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!

Answer 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 $b=x$.

Answer 2

What you said is not correct unless you select a branch. By definition, $log_{\theta}(-1):=ln|-1|+iarg_{\theta} (-1)= 0+iarg_{\theta}(-1)$ So you need $arg_{\theta}(-1)= \pi$. This is true for only one value $\theta$ (so that the branch goes from $\theta$ to $\theta + 2\pi^{-}$). So it seems like the branch with $ -\pi <arg\leq \pi$ should do it.

EDIT What you ultimately need in a branch log is to have the relation $$e^{logz}=z $$ , so that whatever argument $\theta$ you assign to z , you have $$ e^{i\theta}=cos(\theta)+isin(\theta)= i\pi$$. It then follows that $sin\theta= \pi$, so that $\theta =sin^{-1}(\pi)$

Answer 3

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

Answer 4

$e^{\pi i}=-1$ so $e^{\pi i}=i^2$ and $\ln(e^{\pi i})=\pi i$ and $\ln(i^2)=2\ln(i)$

Then $\pi i=2\ln(i)$. Note that $\pi=\frac{2\ln(i)}{i}$ and $e=i^{\frac1{\ln(i)}}$

$\ln(-x)=\ln(xi^2)$ so $\ln(-x)=\ln(x)+2\ln(i)$ or $\ln(-x)=\ln(x)+\pi i$