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Question

Let ω be a cube root of unity not equal to 1. Then the maximum possible value of |a+bω+cω2| where a, b, c ϵ {+1,-1} is

A
0
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B
2
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C
3
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D
1+3
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Solution

The correct option is A 2
ω3=1
ω31=0 or (ω1)(ω2+ω+1)=0
Since ω1,ω=1+i32,ω2=1i32
|a+bω+cω2|=∣ ∣a(b+c)2+i(32)(bc)∣ ∣
Square of the modulus is given by (2abc2)2+(3(bc)2)2
This can be maximised by substituting a=1,b=1,c=1 or a=1,b=1,c=1
Both the substitutions yield the same result, of 4.
maximum modulus =2

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