Question

For integral x,y,z, find the range of |x+yω+zω^2| if it is not true that x=y=z.

(a) [1, ∞)

(b) [√3, ∞)

(c) (0, √3)

(d) (0, ∞)

I have been asked this question during a job interview.

My question is from Roots of Complex Numbers in division Complex Numbers of Engineering Mathematics

Answer 1

Right option is (a) [1, ∞)

Easy explanation: Put a=∞ and b=c=0 to show that maximum value tends to infinity.

Now, z=|x+yω+zω^2|, hence, z^2=|x+yω+zω^2|^2=(x^2+y^2+z^2-xy-yz-zx)=1/2{(x-y)^2+(y-z)^2+(z-x)^2}

Now if x=y, then y≠z and x≠z (given that x,y,z are not all equal)⇒(y-z)^2≥1, also (z-x)^2≥1 and

(x-y)^2=0. Therefore, z^2≥½(0+1+1)=1, hence |z|≥1.