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
|z1z2zm1|ez11|++|zm11
Can anybody tell me how to prove the following inequality?
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Answers (2)

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 |z11|=α1 and |z21|=α2. Then |z1z21|(1+α1)(1+α2)1 Proof: We know α1α2=|z1z2z1z2+1|. By triangle inequality on the three points z1z2, z1+z21, and 1, we have |z1z21||z1z2z1z2+1|+|z1+z22| α1α2+α1+α2 =(1+α1)(1+α2)1 Now, for several numbers, |z1z2zm1|(1+α1)(1+α2,,m)1 (1+α1)(1+α2)(1+α3,,m)1 (1+α1)(1+α2)(1+αm)1 where a2,,m, for example, is my hopefully transparent abuse of notation to denote |z2zm1|. Finally, since 1+xex for real x, the desired inequality follows.
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Vasquez
Answered 2022-01-19 Author has 460 answers
EDIT: This answer is wrong. It all boils down to the inequality |xy1||x1|+|y1|, which I expect to be true. Given this inequality, prove by induction that |z1zm1||z11|++|zm1| Now use ex1+x
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