 # Complex-number inequality |z_{1}z_{2}\cdots z_{m}-1|\leq e^{z_{1}-1|+\cdots+|z_{m}-1|}-1 Can anybody tell me how to prove 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?
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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 $|{z}_{1}{z}_{2}-1|\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)-1$ Proof: We know ${\alpha }_{1}{\alpha }_{2}=|{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|$. By triangle inequality on the three points , and 1, we have $|{z}_{1}{z}_{2}-1|\le |{z}_{1}{z}_{2}-{z}_{1}-{z}_{2}+1|+|{z}_{1}+{z}_{2}-2|$ $\le {\alpha }_{1}{\alpha }_{2}+{\alpha }_{1}+{\alpha }_{2}$ $=\left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)-1$ Now, for several numbers, $|{z}_{1}{z}_{2}\cdots {z}_{m}-1|\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2,\cdots ,m}\right)-1$ $\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)\left(1+{\alpha }_{3,\cdots ,m}\right)-1$ $⋮$ $\le \left(1+{\alpha }_{1}\right)\left(1+{\alpha }_{2}\right)\cdots \left(1+{\alpha }_{m}\right)-1$ 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.
###### Not exactly what you’re looking for? Vasquez
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 ${e}^{x}\ge 1+x$