Complex-number inequality

$|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le {e}^{{z}_{1}-1|+\cdots +|{z}_{m}-1\mid}-1$

Can anybody tell me how to prove the following inequality?

Can anybody tell me how to prove the following inequality?

2022-01-17
Answered

Complex-number inequality

$|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le {e}^{{z}_{1}-1|+\cdots +|{z}_{m}-1\mid}-1$

Can anybody tell me how to prove the following inequality?

Can anybody tell me how to prove the following inequality?

You can still ask an expert for help

nick1337

Answered 2022-01-19
Author has **510** answers

Step 1
The inequality in question bounds how far you can get from 1 by multiplying several complex numbers that may individually not be far from 1. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction.
Lemma: Suppose $|{z}_{1}-1|={\alpha}_{1}$ and $|{z}_{2}-1|={\alpha}_{2}$ . Then $|{z}_{1}{z}_{2}-1|\le (1+{\alpha}_{1})(1+{\alpha}_{2})-1$
Proof: We know ${\alpha}_{1}{\alpha}_{2}=|{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|$ .
By triangle inequality on the three points ${z}_{1}{z}_{2},\text{}{z}_{1}+{z}_{2}-1$ , and 1, we have
$|{z}_{1}{z}_{2}-1|\le |{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|+|{z}_{1}+{z}_{2}-2|$
$\le {\alpha}_{1}{\alpha}_{2}+{\alpha}_{1}+{\alpha}_{2}$
$=(1+{\alpha}_{1})(1+{\alpha}_{2})-1$
Now, for several numbers,
$|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le (1+{\alpha}_{1})(1+{\alpha}_{2,\cdots ,m})-1$
$\le (1+{\alpha}_{1})(1+{\alpha}_{2})(1+{\alpha}_{3,\cdots ,m})-1$
$\vdots $
$\le (1+{\alpha}_{1})(1+{\alpha}_{2})\cdots (1+{\alpha}_{m})-1$
where ${a}_{2,\cdots ,m}$ , for example, is my hopefully transparent abuse of notation to denote $|{z}_{2}\cdots {z}_{m}-1|$ . Finally, since $1+x\le {e}^{x}$ for real x, the desired inequality follows.

Vasquez

Answered 2022-01-19
Author has **460** answers

EDIT: This answer is wrong.
It all boils down to the inequality
$|xy-1|\le |x-1|+|y-1|$ , which I expect to be true.
Given this inequality, prove by induction that
$|{z}_{1}\cdots {z}_{m}-1|\le |{z}_{1}-1|+\cdots +|{z}_{m}-1|$
Now use ${e}^{x}\ge 1+x$

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