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Complex Powers

Date: 09/28/2001 at 03:00:15
From: Bill Koppelman
Subject: Complex powers

How do I show that abs(z^i) < exp^pi where z is a complex number not
equal to 0?

Date: 09/28/2001 at 11:09:29
From: Doctor Mitteldorf
Subject: Re: Complex powers

Start by writing z as r*exp(it) where r and t are real, -pi<t<pi.  

   z^i = r^i * exp(-t)

The first part is

   r^i = exp(i*ln(r)) = cos(ln(r)) + i sin(ln(r))

This has a magnitude of 1. (Any real number raised to the i power has 
magnitude 1.)

The second part is exp(-t), where t is between -pi and pi. The 
greatest value this can have is exp(-(-pi)) = exp(pi).

So there's your proof.  Are you convinced?  

I'm not.  Suppose I had begun by

...writing z as r*exp(it) where r and t are real, 0 < t < 2pi.  

Then I would have finished by saying that the range of exp(-t) was 
exp(-2pi) to exp(0), so the proof would now say abs(z^i) < 1.

Or I might have used the range 100pi < t < 102pi, 

in which case I could have shown that abs(z^i) < exp(-100).

What goes on here? I encourage you to let this paradox rumble around 
in your head, and write to me again with your thoughts about how to 
resolve it. Meanwhile, take a look at this page from our archives:

   e^(i*pi) = -1: pi = 0 ?   

- Doctor Mitteldorf, The Math Forum   
Associated Topics:
High School Imaginary/Complex Numbers
High School Number Theory
High School Transcendental Numbers

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