Complex Cube Roots of Unity and SimplifyingDate: 05/17/2005 at 07:11:54 From: Tom Subject: The complex cube roots of unity With w denoting either of the two complex cube roots of unity, find 2w + 1 2w^2 + 1 -------------- + -------------- 5 + 3w + w^2 5 + w + 3w^2 giving your answer as a fraction a/b, where a, b are integers with no factor in common. I understand you must cross multiply the two fractions but I get stuck when trying to simplify the expression down to the a/b fraction. I cross multiplied and got: 2w^4 + 12w^3 + 16w^2 + 14w + 10 ------------------------------- 3w^4 + 10w^3 + 23w^2 + 20w + 25 But now I can't work out how to simplify. I thought it might have something to do with w^2 * w = 1, but I'm unsure. Date: 05/17/2005 at 12:06:15 From: Doctor Douglas Subject: Re: The complex cube roots of unity Hi Tom. Good work so far. You can keep going because you know that w^3 = 1 and w^4 = w^3*w = 1*w = w, so that the denominator of your fraction simplifies to 3w + 10 + 23w^2 + 20w + 25 = 23w^2 + 23w + 35. The numerator will simplify in a similar way, but you will still end up with a fraction in terms of w^2 and w (and w^0). Here is a very useful fact: 1 + w + w^2 = 0. The three cube roots of unity sum to zero. This is actually true for any set of the Nth roots of unity, not just N=3. You can "prove" this to yourself geometrically, reasoning from the vector sum of the equally-spaced points around the unit circle. To convince yourself analytically, it's a relatively straightforward exercise if you use polar coordinates (sum of a geometric series), but it's a real mess trying to do it in cartesian coordinates. Anyway, you can use this fact to simplify your result. You can also apply this hint to both of the original fractions, e.g., 2w + 1 2w + 1 1 + 2w -------------- = ---------------------- = ------- 5 + 3w + w^2 4 + 2w + (1 + w + w^2) 4 + 2w which is a slightly simpler form. You still have to do some cross-multiplying and use the fact that w^3 = 1, etc., but I think you'll find this to be a little less tedious. You should find that the answer to the original problem is a real number. This is reassuring, because the second fraction is in fact the complex conjugate of the first fraction conjugate(w) = w^2 and conjugate(w^2) = w, so that you have a sum of the form A + conjugate(A), and the result of this sum is 2*Re(A), a real number. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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