Bound difference of complex numbers to bound their log ratio

Question

I'm trying to prove the following claim: if $\delta > 0$ and $a,b \in \mathbb{C}$, then $|a-b| \leq |\frac{a}{e^\delta} - a|$ implies $|\ln(\frac{b}{a})| \leq \delta$. I'm quite certain this is true. I ran a Python script to search for counterexamples with various parameters and found none. I even proved that it holds when $a,b \in \mathbb{R}$. But my complex analysis background is weak, so I need help with the proof for $a,b \in \mathbb{C}$.

Answer 1

We assume that $a \ne 0$ (which implies $ b \ne 0$) and then calling $z=b/a$ the problem becomes $|1-z| \le 1-e^{-\delta}$ implies $|\log z| \le \delta $

Here we note that the inequality $|1-z| \le 1-e^{-\delta} < 1$ implies $\Re z >0$ so we have a canonical logarithm $\log z= \log |z| + i \arg z , |\arg z| < \frac{\pi}{2}$ and of course the inequality is not generally true for other branches as for those the argument, hence |Log $z$|, can go to infinity, so we assume that we use the principal branch with its standard Taylor series.

Then letting $w=1-z, |w| \le 1-e^{-\delta}=c$, $\log z=\log (1-w)=-\sum_{n \ge 1}\frac{w^n}{n}$

By the triangle inequality:

$|\log z| =|\log (1-w)| \le \sum_{n \ge 1}|\frac{w^n}{n}| \le \sum_{n \ge 1}\frac{c^n}{n}=|\log (1-c)|=-\log (1-c)=\delta$ so we are done!