Let ω ≠ 1 be a cube root of unity. Then the minimum of the set { a + b ω + c ω 2 2 : a …

Let ω1 be a cube root of unity. Then the minimum of the set {a+bω+cω22:a,b,c distinct nonzero integers } equals ________

Solution

If z is a complex number
Then |z | 2 =z z ¯
Now,
a+bω+cω22=a+bω+cω2a+bω+cω2¯
a+bω+cω22=a+bω+cω2a+bω-+cω-2
ifaRa-=a
z1z2¯=z-1z-2
ω-=ω2,  ω-2=ω


a+bω+cω22=a+bω+cω2a+bω2+cω
=a2+b2+c2-ab-bc-ca
∵  ω3=1,  1+ω+ω2=0
|a+bω+cω2|2=12[a-b2+b-c2+c-a2]
Now for this to be minimum
Take a=1, b=2, c=3, as a, b, c are distinct and non-zero integers
Minimum of a+bω+cω22=3

Asked in: JEE Advanced 2019 (Paper 1)

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