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Author: rev2217 Big gold star, 5000 posts Old School Fool Add to my Favorite Fools Ignore this person (you won't see their posts anymore) Number: of 5814  
Subject: Re: A (complex) puzzle Date: 9/27/2011 12:40 AM
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Lauren,

Well, no. It is not a precedence problem, because with real numbers the following identity should hold:

(x^a)^b) = x^(ab) = x^(ba) = (x^b)^a

However, if x < 0 then some of the steps may yield complex numbers, and then powers and roots (and logarithms, etc) are operations that can have multiple answers. None of those answers are wrong, so it is not a precedence issue. In fact, if you were to define a "required" precedence, then you would be eliminating valid answers.


I hate to break this to you, but it absolutely is a problem of precedence. Your supposed identity is valid only if x is a positive real value or zero -- and you actually provided a counterexample that shows how it fails when x is negative.

Note that considering all four fourth roots implied by the exponent of 3/4, which differ by multiples of 90 deg. in phase angle, also does not solve this problem. If one evaluates all four roots of the original expression, one gets 8, 8i, -8, and -8i. Exchanging the exponents, the four roots of (-4)^(3/4) have phase angles of 45, 135, -135, and -45 degrees. When squared, these values yield only 8i and -8i -- one does not get the real values.

Norm.
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