Complex PowersDate: 04/10/2002 at 16:43:02 From: Brian England Subject: Complex exponents I recently came across an interesting result and was wondering if you could help. Given e^(2*pi*i/2*pi*i) = e^(1) = e but e^(2*pi*i/2*pi*i) = (e^(2*pi*i))^(1/2*pi*i) and we know e^(2*pi*i) = 1 so 1^(1/2*pi*i) has to be equal to e. I am having trouble proving this last step. I have rewritten the equation many times, but keep getting stuck. Thanks, Brian England Date: 04/10/2002 at 22:48:28 From: Doctor Peterson Subject: Re: Complex exponents Hi, Brian. Part of the problem here is that complex powers are multi-valued; there is not just one value of i^i, for example (though we often give one answer, it is really just the principal value, much as we give one square root of 2 when there are really two of them). For example, since 1 = e^(2 k pi i) for any integer k we can say that 1^i = e^(2 k pi i i) = e^(-2 k pi) Of these infinitely many values, the principal one is for k = 0, namely 1 (of course). What's interesting here is that, whereas we say that 1 to any power is 1, for complex exponents that is just one of the possible values. Your case is really just this raised to the -1/(2 pi) power: 1^(1/(2 pi i)) = 1^(-i/(2 pi)) = (1^i)^(-1/(2 pi)) = e^(-2k pi * -1/(2 pi)) = e^k FOR ANY INTEGER k. In other words, although, as you saw, your expression gives e, it also gives the expected 1 (for k = 0) and all other powers of e. I don't recall ever applying this idea to powers of 1 before. Maybe I just couldn't bear the weirdness of it all! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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