The bound of a log function It looks like we can control $\log\frac{1+z}{1-z}$ by $\log\frac{1+r}{1-r}$ if $|z|=r<1$ where the logarithm is defined on the branch obtained by deleting the negative imaginary axis. 
I tried to prove it but it's not so easy for me. 
Somebody can help me to solve this? Thanks.

Seeing that my attempt to mark this question as a duplicate failed, I copy the answer from the other thread: 
$$\left|\log\frac{1+z}{1-z}\right|=2\left|\int_0^z \frac{d\zeta}{1-\zeta^2}\right| \le 
2 \int_0^{|z|} \frac{dt}{1-t^2} =\log\frac{1+|z|}{1-|z|}$$
