Fractional Exponents and Complex RootsDate: 06/11/99 at 03:59:08 From: David Hanson Subject: Complex roots We are having a discussion about unique representations of complex roots. Does z^(a/b) = (z^a)^(1/b) or (z^1/b)^a, and does z^(1/d) represent the complex number with the smallest positive argument, or is it ambiguous notation for all the d roots? For example, does [(-4)^2]^(1/4) = 16^(1/4) = the principal fourth root of 16, which is 2, but [(-4)^(1/4)]^2, where if you take the fourth root of -4 with the smallest positive argument and then square it, you get 2i? Can you help? Thanks. David Hanson Date: 06/11/99 at 05:15:44 From: Doctor Mitteldorf Subject: Re: Complex roots Dear David, Even simpler, you can take (-2), square it, find the square root, and end up with a different number than you started with. My advice is to keep your eye on the ball. The conventions are there to serve you, not to confine your thinking, and besides, conventions in this area are murky. You and your students should be aware of this kind of ambiguity, and alerted that when you are trying to find specific solutions to equations, ambiguity may arise so you will need to examine all roots, then think about what makes sense in the context of the present problem. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 06/11/99 at 05:42:58 From: Doctor Pete Subject: Re: Complex roots Hi, As you can see from your examples, it is generally incorrect to take "principal" values during intermediate steps in evaluation. Doing so is equivalent to losing solutions, and the result is that the normal axioms of associativity, commutativity, and cancellation may be violated. If we are to make any sense of z^(a/b) then it is evident that one must be careful in saying what a "b-th" root of a complex number is. Such a root is not a number, but a set. It is simply not correct to say that 4^(1/2) = 2 rather, when dealing with the complex numbers, 4^(1/2) = {2, -2}. Similarly, 1^(1/3) = {1, e^(2*pi*i/3), e^(-2*pi*i/3)}. When dealing with complex numbers, one must become accustomed to the idea that functions that are bijective (one-to-one and onto) in the reals are not necessarily so in the complex plane, and hence functions should not be thought of as mappings of a number to a number, but of a set to a set. In the cube root of unity example above, the function f(z) = z^(1/3) is a one-to-three mapping, taking (nearly every) complex number z and mapping it to three distinct elements. Similarly, the inverse function f^(-1)(z) = z^3 is a three-to-one mapping, where for a given value of z^3, there are three complex numbers that map to z^3 (one of them being z itself). So it is at once clear that z^(1/n) is not "ambiguous" in the sense that its value is not specified - what is *meant* by the symbol z^(1/n) is not "the" n-th root of z, but the "n" n-th roots of z. A set of values is no more ambiguous than a single value. To demonstrate how this changes your example, consider (-4)^(2/4). Apply the axiom of cancellation to obtain (-4)^(1/2) = {2i, -2i}. This is the correct solution set. On the other hand, one could do it this way: (-4)^(1/4) = {1+i, -1+i, -1-i, 1-i} so [(-4)^(1/4)]^2 = {(1+i)^2, (-1+i)^2, (-1-i)^2, (1-i)^2} = {2i, -2i, 2i, -2i} = {2i, -2i}. But why isn't [(-4)^2]^(1/4) = (16)^(1/4) = {2, 2i, -2, -2i} valid? This is for the same reason that -1 = i^2 = (-1)^(1/2)(-1)^(1/2) = [(-1)(-1)]^(1/2) = 1^(1/2) = 1 is invalid. What's going on is that we're mapping 2-to-1, then 1-to-2, creating extraneous solutions. Since this creates contradictions such as the above, it follows that whenever possible, we first evaluate one-to-many, then many-to-one. So the answer to your original question is that z^(a/b) indicates [z^(1/b)]^a. But this consideration arises only in the case where a and b have a common factor. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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