Quick modulus question - complex exponential function

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Discussion Overview

The discussion revolves around the properties of the complex exponential function, specifically the expression |exp(z^n)| when the modulus of z is less than 1. Participants explore the implications of this condition within the context of complex analysis.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant questions what |exp(z^n)| is less than when |z| < 1, suggesting it might be e.
  • Another participant proposes using the binomial expansion of (a + ib)^n to analyze the expression further.
  • A later reply confirms the initial thought, providing a mathematical inequality: |exp(z^n)| = exp(Re(z^n)) ≤ exp(|z^n|) = exp(|z|^n) < exp(1^n) = e, indicating that |exp(z^n)| is indeed less than e under the given condition.
  • One participant expresses relief at finding agreement with their initial thought process, despite feeling mentally fatigued from studying complex analysis.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical reasoning that |exp(z^n)| is less than e when |z| < 1, though the discussion includes exploratory reasoning and some uncertainty in articulating the concepts.

Contextual Notes

The discussion does not resolve potential nuances in the application of the binomial expansion or the implications of the inequality presented.

buckylomax
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What is |exp (z^n)| less than if |z| < 1? I'm thinking it's e but I'm having a brain freeze at the moment! Thanks for any help guys.
 
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put z=a+ib then expand $$(a+ib)^n$$ using the binomial expansion ...
 
buckylomax said:
What is |exp (z^n)| less than if |z| < 1? I'm thinking it's e but I'm having a brain freeze at the moment! Thanks for any help guys.

Welcome to MHB, buckylomax! :)

That sounds fine to me.

$$|\exp(z^n)| = \exp(\Re(z^n)) \le \exp(|z^n|) = \exp(|z|^n) < \exp(1^n) = e$$
 
I like Serena said:
Welcome to MHB, buckylomax! :)

That sounds fine to me.

$$|\exp(z^n)| = \exp(\Re(z^n)) \le \exp(|z^n|) = \exp(|z|^n) < \exp(1^n) = e$$

That's what I was thinking but I just couldn't articulate it to the end. I think I broke my brain because I've been studying complex analysis for the past 8 hours (Puke)

Thanks a lot for the help.
 
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