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

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
