Is it always true that $Log(z^{-1})=-Log(z)$

Question

Is it always true that $Log(z^{-1})=-Log(z)$?

Can we just write $z=re^{i\theta}$ and then

$$-Log(z)=-(\ln r+i\theta)=-\ln r-i\theta=\ln (r^{-1})+i(-\theta)=Log(r^{-1}e^{-i\theta})=Log(z^{-1})$$

it seems suspicious to me.

Note:

For $z\in\mathbb{C}\setminus (-\infty,0]:Log(z)=ln|z|+iArg(z)$

Clarification:

I know that it is not always true that $Log(z^{2})=2Log(z)$.

So, writing $z=re^{i\theta}$ and

$$2Log(z)=2(\ln r+i\theta)=2\ln r+2i\theta=\ln (r^{2})+i(2\theta)=Log(r^{2}e^{2i\theta})=Log(z^{2})$$

is false. And to me it looks kind of the same (maybe false) "technique".

Answer 1

If you restrict $z$ to $\mathbb{C}\setminus (-\infty,0]$ and choose the argument such that $-\pi < Arg(z) < \pi$ then it is true:

If $z = r e^{i\theta}$ with $-\pi < \theta < \pi$ then $z^{-1} = r^{-1} e^{-i\theta}$ and $-\pi < -\theta < \pi$, therefore $$ Arg(z^{-1}) = - \theta = -Arg(z) $$ which implies $$ Log(z^{-1}) = \ln |z^{-1}| + i Arg(z^{-1}) = - \ln |z| - i Arg(z) = - Log(z) \, . $$

This argument fails with $Log(z^2)$ because $-\pi < \theta < \pi$ does not imply $-\pi < 2\theta < \pi$.