If $|z|<1 $ then $Log(1-z^2)=Log(1-z)+Log(1+z)$ , $ z\in\mathbb{C}$ Verify that $Log(1-z^2)=Log(1-z)+Log(1+z)$ when  $|z|<1$. 
What can be said about $Log(\frac{1-z}{1+z})$ for such $z$ ?
I tried . But no conclusion.
Here $Log$ is the principal branch.
 A: without using series
$\text{Log}(1-z)+\text{Log}(1+z)=\ln|1+z|+i\text{Arg}(1+z)+\ln|1-z|+i\text{Arg}(1-z)=\ln|(1+z)(1-z)|+i(\text{Arg}(1+z)+\text{Arg}(1-z))=\ln|1-z^2|+i\text{Arg}(1-z^2)=\text{Log}(1-z^2)$
A: The Taylor series of $\log{(1-z^2)}, \:\log{(1-z)},\:\log{(1+z)}$ are absolute convergent in $|z|<1$. So
$$
\log{(1-z)}+\log{(1+z)}=-\sum_{n=1}^{\infty}\frac{z^{n}}{n}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{z^{n}}{n}=-\sum_{n=1}^{\infty}\frac{z^{2n}}{n}=\log{(1-z^2)}
$$
A: This is an exercise from Conway's "Functions of one complex variable" (Ch. III, Sec. 2, Ex. 20) that gives a sufficient condition for the log of a
product to be the sum of the logs.
Suppose $z_k$ are such that $\operatorname{re} z_1\cdots z_i >0$ for $i=1,...,n$. Then
$ \operatorname{Log} (z_1\cdots z_n) = \operatorname{Log} z_1 + \cdots + \operatorname{Log} z_n$. This is straightforward to establish using the
polar form of $z_k$.
In the above example, if $|z|<1$, then $\operatorname{re} (1-z) >0$,
$\operatorname{re} (1+z) >0$, $\operatorname{re} {1 \over 1+z} >0$, $\operatorname{re} {1-z \over 1+z} >0$
and $\operatorname{re} (1-z^2) >0$.
