Determine all solutions to $\operatorname{Log}(z^2) = \operatorname{Log}(z)$ (Complex Equation)

Question

I am new to complex analysis and i was trying to solve the given exercise

Determine all solutions to $\operatorname{Log}(z^2) = \operatorname{Log}(z)$

However, i am not sure whether my attempt is correct or complete.

My attempt:

It holds $$\operatorname{Log}(z^2) = \ln( \vert z \vert^2) + i\operatorname{Arg}(z^2) = 2\ln(\vert z \vert) + 2i\operatorname{Arg}(z) $$

therefore

$$\operatorname{Log}(z^2) = \operatorname{Log}(z)$$ is true for all $z \in \mathbb{C}$ for which

$$2\ln(\vert z \vert) + 2i\operatorname{Arg}(z) = \ln(\vert z \vert) + i\operatorname{Arg}(z)$$ $$\Leftrightarrow \ln(\vert z \vert) + i\operatorname{Arg}(z) = 0$$

This is how far i got. Can i somehow solve this equation?

Thanks for any hints!

Answer 1

Just put together the two conditions you have got right there and you're done.

The $\ln$ term is the real part of that last expression, and $i\text{Arg}$ is the imaginary part, so they must both be $0$. $\ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i \text{Arg} z = 0 \implies \text{Arg} z = 0$, which means that $z$ is actually a positive real number.

The only $z$ satisfying both conditions is $z = 1$.

Answer 2

If $z \ne 0$ and $\operatorname{Log}(z^2) = \operatorname{Log}(z)$, then

$z^2=e^{\operatorname{Log}(z^2)}=e^{\operatorname{Log}(z)}=z$, hence $z=1$.