This is a question from complex numbers.

Question

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 ?

Answer 1

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 ω³ = 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 

 

ω² = e^i4π/3 

 

is a complex number, which, when cubed, gives us 1, and similarly,

 

ω³ = 1.

 

In other words, both ω and ω² 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 have

 

a = q

b = qω

c = qω².

 

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 + zqω²)/-(xqω + yqω² + zq)

                                            = (x + yω + zω²)/(xω + yω² + z)

 

If we now multiply numerator and denominator by ω², and use the fact that ω⁴ = ω (which you can verify), we will have

 

ω²(x + yω + zω²)/(x + yω + zω²) = ω²

 

as claimed.

 

Regards,

Hassan