Is $\log(-1)$ equal to $-\log(-1)$

Question

I thought it should be because if the logarithmic identities hold then, $$-\log(-1)=\log(-1^{-1})=\log(-1)$$ But $\log(-1)=i*\pi$ and $-\log(-1)=-i*\pi$

Answer 1

Logarithmic identities that you learnt hold for "positive" numbers only.

Answer 2

The principal branch of the logarithm is $\log(-1)=\pi i$, which doesn't satisfy the identity you're looking for: $-\pi i\neq\pi i$.

The multi-valued logarithm is $\mathrm{Log}(-1)=\pi i + 2\pi i\mathbb Z$, which does satisfy the identity: $$-(\pi i + 2\pi i\mathbb Z)=\pi i + 2\pi i\mathbb Z.$$

Answer 3

In complex numbers, for a complex number $z=re^{i\theta}$, $$ \ln(z)=\ln(r)+i(\theta+2\pi k) $$ where $k$ is any integer. (This is called a multi-valued function because there is more than one output for a given input, namely all the possible values for a given $k$.)

In your situation, $r=1$ and $\theta=\pi$. But note that $-\pi$ and $\pi$ differ by $2\pi$, so you're just getting two values that are part of the same multi-valued output. So you get $i(\pi+2\pi 0)=i\pi$ and $i(\pi+2\pi(-1))=-i\pi$.