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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.

How do I proceed now? Please help me in this regard.

Thanks a lot.

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

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  • $\begingroup$ Good point @Kavi Rama Murthy. $\endgroup$ – A.Chattopadhyay May 13 at 6:57

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