    Question

# Let z1,z2,z3,z4,z5 and z6 be complex numbers lying on a unit circle with centre (0,0). If ω=(6∑k=1zk)(6∑k=11zk), then

A
ω is purely real.
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B
ω is purely imaginary.
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C
Maximum possible value of |ω| is 6.
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D
Maximum possible value of |ω| is 36.
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Solution

## The correct option is D Maximum possible value of |ω| is 36.|zk|=1⇒1zk=¯¯¯¯¯zk ω=(6∑k=1zk)(6∑k=11zk)ω=(z1+z2+⋯+z6)(1z1+1z2+⋯+1z6)=(z1+z2+⋯+z6)(¯¯¯¯¯z1+¯¯¯¯¯z2+⋯+¯¯¯¯¯z6)=(z1+z2+⋯+z6)(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2+⋯+z6) ω=|z1+z2+⋯+z6|2 which is purely real. 0≤|z1+z2+⋯+z6|≤|z1|+|z2|+⋯+|z6|⇒0≤(|z1+z2+⋯+z6|)2≤(|z1|+|z2|+⋯+|z6|)2⇒0≤|ω|≤36  Suggest Corrections  0      Similar questions
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