Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Complex Cube Roots of Unity and Simplifying

Date: 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/ 
Associated Topics:
College Imaginary/Complex Numbers

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/