# A logarithmic inequality.

Show that $$|\log (1-z)| \leq |z| + \frac {|z|^2} {(1-|z|)^2},$$ for all $$z$$ with $$|z|<1.$$

I know that $$\log(1-z) = \log |1-z| + i\ \text {arg} (1-z).$$ This shows that \begin{align*} |\log(1-z)| = \sqrt {(\log|1-z|)^2 + (\text {arg}(1-z))^2} &\leq \sqrt {(|1-z|^2-1)^2 + {\pi}^2}.\\ & =\sqrt {\left (|z|^2-2\ \mathfrak {R} (z) \right )^2 + {\pi}^2}.\\ &\leq \sqrt {|z|^4-2\ |z|^2\ \mathfrak {R} (z) + 4\ {\mathfrak {R} (z)}^2 + {\pi}^2}. \end{align*} for the principal branch of logarithm.

• Did you try to use the power series for $\log(1-z)$? – Kavi Rama Murthy May 13 at 6:42
$$|\log(1-z)|\leq |z|+|z|^{2}/2+|z|^{3}/3+... \leq |z| +|z|^{2} (1+|z|+|z|^{2}+...)=|z| +|z|^{2}/(1-|z|)$$. This is even stronger than your inequality.