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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 $(-\pi,\pi].$

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What's your definition of $\text{Log}$? A power series? Then you need $|z| < 1$ to ensure absolute convergence. – Qiaochu Yuan Mar 6 '12 at 0:06
@QiaochuYuan: $\operatorname{Log(z)}=\ln|z|+i\operatorname{Arg(z)}$. – spohreis Mar 6 '12 at 0:15
And what's your definition of $\text{Arg}$? – Qiaochu Yuan Mar 6 '12 at 0:17
It is the unique argument of $z$ that belongs to $(-\pi,\pi]$. – spohreis Mar 6 '12 at 0:25
up vote 2 down vote accepted

I assume they are using $\text{Arg}$ for the principal branch of the argument, i.e. $-\pi < \text{Arg}\ w \le \pi$. You'll have to check that $-\pi < \text{Arg}\ (1-z) + \text{Arg}\ (1+z) \le \pi$. Hint: if $1-z$ is in the upper half plane, $1+z$ is in the lower, and vice versa. The result is actually true for all $z \ne \pm 1$.

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Both sides of the equations are holomorphic functions in the unit disk, and they are equal on the interval $(-1,1)$. By the identity theorem, they must be equal everywhere in the disk.

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Perfect job. Simple and elegant. – luka5z Nov 25 '15 at 10:48

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