I am trying to prove the following inequality:
$|Log(1+z)|\leq -\log(1-|z|)$ for $|z|<1$
where Log is the principal branch and log is the natural logarithm of the positive number $1-|z|$.
So far I have,
$|Log(1+z)|=\bigg|\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}z^n\bigg|\leq \sum\limits_{n=1}^{\infty}\frac{|z|^n}{n}$
and I also know that for $|z|<1$,
$-Log(1-z)=\sum\limits_{n=1}^{\infty}\frac{z^n}{n}$.
Is the last equation valid if I replace $z$ with $|z|$ and $Log$ with $\log$? I usually prove these Log power series by taking derivatives/antiderivatives, but I know that $|z|$ is not complex differentiable.
Yes, your last equation is valid if you substitute $z$ for $|z|$, as you are simply evaluating your function, and then you have $\mathrm{Log}(1 - |z|) = \log(1 - |z|)$ so you can indeed replace $\mathrm{Log}$ by $\log$ (but you needed to take these actions in that order of course).
You do not have to prove the substitution, as you're merely using a pre-established formula and using it on some select numbers.