Non-power series proof that $|\log (1 + z)| \le -\log (1 - |z|)$

Question

Given $z \in \mathbb{C}$, $|z| < 1$, we have the bound

\begin{align*} |\log (1 + z)| &= \left| \sum_{n \ge 1} \frac{(-1)^{n+1}z^n}{n} \right| \\ &\le \sum_{n \ge 1} \frac{|z|^n}{n} \\ &= -\sum_{n \ge 1} \frac{(-1)^{n+1}(-|z|)^n}{n} \\ &= -\log (1 - |z|) \end{align*}

That is, the largest magnitude of $\log (1 + z)$ within a given radius $r$ of $z = 0$ is achieved at $z = -r$.

Question: Is there a simple derivation of this bound that uses only "algebraic" or other simple properties of $\log z$, and not the power series?

I'm being slightly vague about what properties are allowed, but in particular I want to avoid infinite series. (The motivation is a proof that's easy to formalize as part of https://github.com/girving/ray; I could do the series proof but it seems like there should be something simpler.)

Answer 1

Assume that $|z|<1$. With the change of integration variables $t=1+zs$, we find that $$ \log (1 + z) = \int_1^{1 + z} {\frac{{{\rm d}t}}{t}} = z\int_0^1 {\frac{{{\rm d}s}}{{1 + zs}}} . $$ Then $$ \left| {\log (1 + z)} \right| \le \left| z \right|\int_0^1 {\frac{{{\rm d}s}}{{1 - \left| z \right|s}}} = - \log (1 - \left| z \right|). $$