Question

Let ω and ω^2 be the non-real cube roots of unity and 1/(a+ω)+1/(b+ω)+1/(c+ω)=2ω^2 and 1/(a+ω^2)+1/(b+ω^2)+1/(c+ω^2)=2ω, then calculate 1/(a+1)+1/(b+1)+1/(c+1).

(a) 1

(b) 2

(c) 3

(d) 4

I have been asked this question in class test.

This question is from Roots of Complex Numbers topic in section Complex Numbers of Engineering Mathematics

Answer 1

Right option is (b) 2

Easy explanation: Given relations can be written as: 1/(a+ω)+1/(b+ω)+1/(c+ω)=2/ω, and

1/(a+ω^2)+1/(b+ω^2)+1/(c+ω^2)=2/ω^2⇒ω and ω^2 are roots of  1/(a+x)+1/(b+x)+1/(c+x)=2/x.

⇒[3x^2+2(a+b+c)x+bc+ca+ab]/[(a+x)(b+x)(c+x)]=2/x⇒x^3+(bc+ca+ab)x-2abc=0.

Let 3^rd root be α (apart from ω and ω^2). Then, α+ ω+ ω^2=coefficient of x^2=0⇒α=1. Hence, 1/(a+1)+1/(b+1)+1/(c+1)=2/1=2.