Complex analysis: find contradiction of a relationship

[A Story of a Student]

Homework Statement

Find a $z \in \mathbb{C}$ such that $log(1/z)\neq −log(z)$

Relevant Equations

$\log(z)=\ln|z|+i\arg(z)$

I have reached a conclusion that no such z can be found. Are there any flaws in my argument? Or are there cases that aren't covered in this?

Attempt
$\log(\frac{1}{z})=\ln\frac{1}{|z|}+i\arg(\frac{1}{z})$
$-\log(z)=-\ln|z|-i\arg(z)$

For the real part $\ln\frac{1}{|z|}=\ln1-\ln|z|=-\ln|z|$
For the imaginary part $\arg(\frac{1}{|z|})=\arg(\frac{1}{z\overline z}\overline{z})=\arg(\frac{1}{|z|^2}\overline{z})=-\arg{z}$

Thus $\log(\frac{1}{z})=-\log(z)$ for $z\in\mathbb{C}\backslash\{0\}$

Mentor note: Fixed the broken LaTeX. Log and Arg should not be capitalized.

 

[WWGD]

How about for $z=i \rightarrow 1/z= -i$?

 

[A Story of a Student]

How about for $z=i \rightarrow 1/z= -i$?

I found they to be same.
$\log(\frac{1}{z})=\log(-i)=\ln(1)+i\arg(-i)=-i\frac{\pi}{2}$
$-\log(z)=-\ln(1)-i\arg(i)=-i\frac{\pi}{2}$

 

[WWGD]

I found they to be same.
$\log(\frac{1}{z})=\log(-i)=\ln(1)+i\arg(-i)=-i\frac{\pi}{2}$
$-\log(z)=-\ln(1)-i\arg(i)=-i\frac{\pi}{2}$

Ok, I may have been wrong, sorry. I think you may have been right. $1/z$ =$\frac { z^{-} }{z^{-}z}$ where
given $z=x+iy, z^{-}= x-iy$. The map $z \rightarrow z^{-}$ reflects $z$ along the $x-$ axis, so the angles are equal with respect to the x-axis , but of different sign. My bad.

 

[FactChecker]

If log(z) is restricted to the principal branch, then can you find a z where the two are not equal within the principal branch?

 

[WWGD]

If log(z) is restricted to the principal branch, then can you find a z where the two are not equal within the principal branch?

Yes, good point, depending on where the cut is this will be possible. But there may be issues on whether the reflection "Jumps the branch".

 

[WWGD]

If log(z) is restricted to the principal branch, then can you find a z where the two are not equal within the principal branch?

I think they will be equal modulo $2k\pi; k \in \mathbb Z$.

 

[FactChecker]

I think they will be equal modulo $2k\pi; k \in \mathbb Z$.

Right. But that is the only way I can see to make the problem correct.