2022-01-17

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?

nick1337

Expert2022-01-19Added 777 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

Expert2022-01-19Added 669 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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