Is $\log(z^2)=2\log(z)$ if $\text{Log}(z_1 z_2)\ne \text{Log}(z_1)+\text{Log}(z_2)$? For any $z1, z2$ in $\mathbb{C} \setminus {0}$, $\log(z_1 z_2)=\log(z_1)+\log(z_2)$, but in general $\text{Log}(z_1 z_2)\ne \text{Log}(z_1)+\text{Log}(z_2)$. 
Is $\log(z^2)=2\log(z)$?
 A: 
The answer to the question in the OP is that in general $\displaystyle \log(z^2)\ne 2\log(z)$. 

This might seem paradoxical given the relationship expressed as
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag1$$
But $(1)$ is interpreted as a set equivalence.  It means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$.  And conversely, it means that the sum of any value of $\log(z_1)$ and any value of $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.
Note that $(1)$ is true since $\log(|z_1z_2|)=\log(|z_1|)+\log(|z_2|)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$. 


EXAMPLE:

As an example, suppose that $z_1=z_2=-1$.  Then, $z_1z_2=1$ and $\log(1)=i2n\pi$ for any integer $n$.  For $n=0$, $\log(1) =0$.  Then $(1)$ is certainly satisfied by $\log(z_1)=i\pi$ and $\log(z_2)=-i\pi$.

Note that although $z_1=z_2$ here, we needed to take two distinct values for $\log(z_1)$ and $\log(z_2)$.  We were afforded that degree of freedom since we viewed $z_1$ and $z_2$ as independent.


In general, 
$$\log(z^2)\ne 2\log(z) \tag 2$$
To see this, we note that for $z=re^{i\theta}$, 
$$2\log(z)=2\log(r)+i2(\theta+2n\pi) \tag 3$$
while
$$\begin{align}
\log(z^2)&=\log(r^2e^{i2\theta+i2n\pi})\\\\
&=2\log(r)+i2(\theta+n\pi)\tag 4
\end{align}$$
Comparing $(3)$ and $(4)$ we see that $\log(z^2)$ and $2\log(z)$ do not share the same set of values.


EXAMPLE:

As an example, suppose $z=i$.  For the value $\log(i^2)=i3\pi$, there is no corresponding value of $2\log(z)$.  Hence, $\log(i^2)\ne 2\log(i)$ in general. 
A: If $p$ is a non-integer then $z^p$ is a complex multi-valued function and the principal value of $\ln z$  must lie in $- \pi < \Im \ln z< \pi$. From this it follows that
\begin{align}
\ln (z_1 z_2) &= \ln z_1 + \ln z_2 + 2 \pi i N_{+} \\
\ln \left( z_1 / z_2 \right) &= \ln z_1 - \ln z_2 + 2 \pi i N_{-}\\
\ln z^n &= n \ln z + 2 \pi i N_{n}
\end{align}
Where $N_{\pm} = 0, +1, -1$ and
$$N_n = \frac{1}{2}+\left(\frac{n}{2 \pi}\right)\arg z$$
