Find the error in following reason $(-z)^2=z^2 \implies \log(-z)^2=\log(z)^2 \implies2\log(-z)=2\log(z)\implies \log(-z)=\log(z)$ Find the error in following reason
\begin{align*}
(-z)^2=z^2
&\implies \log(-z)^2=\log(z)^2\\
&\implies2\log(-z)=2\log(z)\\
&\implies \log(-z)=\log(z)
\end{align*}
I think the error is $2\log(-z)=2\log(z)$ because for $z=1$, $2\log(1)=0$, but $2 \log(-1)$ is undefined.
What's bugging me is this problem has a star on it, which mean it's a challenging problem. I don't think it is that easy. So I wonder if anyone could check if I missed or make mistake somewhere.
 A: The problem is that $\log z$ is multivalued, and so requires a branch cut.
$\log z=\log|z|+i\,\text{arg}\,z$. We can choose the branch $-\pi<\text{arg}\,z\le\pi$. Now, let $z=e^{\frac{3\pi i}4}$.
$$\log z=\log e^{3\pi i/4}=\frac{3\pi i}4.$$
But 
$$\log z^2=\log(-i)=-\frac{\pi i}2.$$
Edit:
This shows that $\log z^2=2\log z$ does not hold for complex numbers.
A: Let's make it even simpler:  If you know that $(-z)^2 = z^2$, does it necessarily follow that $-z = z$?  Why not?
A: The error is in the second implication.
If we work in the real numbers, we have that $\log x^2=2\log|x|$. This follows from the more general notion of Complex Logarithm, which is the one we use when we work with the complex numbers; it is (briefly) defined as follows (once you have choosen the principal branch which allows to work with the so called Principal Logarithm)
$$
\log z=\log|z|+i\arg z
$$
where $\log|z|$ is the real log.
Clearly $\log|-z|=\log|z|$, but in general $\arg(-z)$ is NOT equal to $\arg z$ (very roughly speaking, if you write $z=re^{i\theta}$ then $\arg z=\theta$).
