MHB Quick modulus question - complex exponential function

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The discussion revolves around determining the upper limit of |exp(z^n)| when |z| < 1. It is established that |exp(z^n)| is less than e, as shown through the inequality |exp(z^n)| = exp(Re(z^n)) ≤ exp(|z^n|) = exp(|z|^n). Since |z| is less than 1, it follows that exp(|z|^n) is also less than e. Participants confirm this reasoning, with one expressing relief at articulating the thought process after an intensive study session. The conclusion reinforces that |exp(z^n)| is indeed less than e for |z| < 1.
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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Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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