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Question

Let z0 be a root of the quadratic equation, x2+x+1=0. If z=3+6iz8103iz930, then argz is equal to :

A
π4
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B
π6
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C
π3
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D
0
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Solution

The correct option is A π4
Roots of x2+x+1=0 are ω and ω2, where ω is the cube root of unity.
Let z0=ω
Thus, z=3+6i(ω)813i(ω)93
=3+6i3i [ω3=1]
=3+3i

arg(z)=tan1(33)=tan1(1)
arg(z)=π4

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