This is Exercise EP$20$ from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa.
Check that $\operatorname{Log}(1-z^{2})=\operatorname{Log}(1-z)+\operatorname{Log}(1+z)$ when $|z|\lt 1$. What one can say about $\operatorname{Log}\frac{(1-z)}{(1+z)}?$
I'll be honest with you. I don't know how to use $|z|\lt 1$. Let me show what I did so far $\operatorname{Log}(1-z^{2})=\ln|1-z^{2}|+i\operatorname{Arg}(1-z^{2})=\ln|1-z|+\ln|1+z|+i\operatorname{Arg}(1-z^{2})$. Should I use the hypothesis $|z|\lt 1$ to prove that $\operatorname{Arg}(1-z^{2})=\operatorname{Arg}(1-z)+\operatorname{Arg}(1+z)?$
EDIT: Where $\operatorname{Log(z)}=\ln|z|+i\operatorname{Arg(z)}$ and $\operatorname{Arg(z)}$ is the unique argument of $z$ that belongs to $(-\pi,\pi].$