Complex PowersDate: 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. Then 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 ? http://mathforum.org/dr.math/problems/koehler10.17.97.html - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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