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 Suppose z is a complex number such that z1|z|=1, and  arg(z)=θLet ω =z(1z¯)z¯(1+z),than Re(ω) ) is equal to

a
1+cos⁡(θ/2)
b
1−sin⁡(θ/2)
c
−2sin2⁡(θ/2)
d
2cos2⁡(θ/2)

detailed solution

Correct option is C

ω=z−zz¯z¯+z¯z=z−11/z+1=(z−1)zz+1 [∵|z|=1]=z2−1z+1−z−1z+1=z−1+1−z1+zRe⁡(ω)=Re⁡(z)−1+Re⁡1−z1+zBut Re⁡1−z1+z=121−z1+z+1−z1+z=121−z1+z+z−1z+1=0∴ Re⁡(ω)=cos⁡θ−1=−2sin2⁡(θ/2)

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