Vaibhav P.

asked • 05/14/16# This is a question from complex numbers.

If 1,w and w2 are the three cube roots of unity and a,b and c are the cube roots of p,q (<0),then for any x,y and z,the expression {(xa + yb + zc)÷(xb + yc + za)} is equal to ?

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## 1 Expert Answer

Hassan H. answered • 05/16/16

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Vaibhav,

Sorry, I forgot about this question over the weekend, as I got pretty busy. Perhaps you already have decoded the solution, but based on what you wrote, I will give you my interpretation of what is going on.

From the solution you present, it seems that p is supposed to be a negative real number, and q perhaps represents the absolute value of its cube roots. We name the three values of the cube roots as a, b, and c.

Now, the symbol ω represents the primary cube root of unity. That means it is the complex number

ω = e

^{i2π/3}and note that ω

^{3}= 1. It is a basic theorem in complex analysis that we can express the nth roots of a complex number in terms of combinations of the nth roots of unity, which are similarly defined.Note also that

ω

^{2}= e^{i4π/3}is a complex number, which, when cubed, gives us 1, and similarly,

ω

^{3}= 1.In other words, both ω and ω

^{2}are cube roots of unity. Since 1 itself is also a cube root of unity (unity here just means 1 of course), we can now represent the cube roots of any real number p in terms of these roots of unity. Letting q = p^{1/3}, we would havea = q

b = qω

c = qω

^{2}.If p < 0, we can of course choose q = -p

^{1/3}, or what is the same, -q = p^{1/3 }and it seems that is what your question asked. In any case, setting up the expression we are asked to evaluate, we now get(xa + yb + zc)/(xb + yc + za) = -(xq + yqω + zqω

^{2})/-(xqω + yqω^{2}+ zq) = (x + yω + zω

^{2})/(xω + yω^{2}+ z)If we now multiply numerator and denominator by ω

^{2}, and use the fact that ω^{4}= ω (which you can verify), we will haveω

^{2}(x + yω + zω^{2})/(x + yω + zω^{2}) = ω^{2}as claimed.

Regards,

Hassan

Vaibhav P.

Hassan,

Thanks,actually I also forgot about the question but now I definitly got what's going on in this question.

Thanks again.

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05/19/16

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Hassan H.

05/14/16