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### 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

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

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