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#### buckylomax

##### New member

- Aug 3, 2012

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- Thread starter buckylomax
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- Aug 3, 2012

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- Jan 17, 2013

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put z=a+ib then expand \(\displaystyle (a+ib)^n\) using the binomial expansion ...

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- Mar 5, 2012

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Welcome to MHB, buckylomax!

That sounds fine to me.

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

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- Aug 3, 2012

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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 hoursWelcome to MHB, buckylomax!

That sounds fine to me.

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

Thanks a lot for the help.

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