Question

# If z=x+iy, |z|=1 and ω=(1−z)21−z2, then the locus of ω is equivalent to |z−2|=|z+2||z−2i|=|z+2i||z−1−i|=|z+1−i||z−i|=|z+i|

Solution

## The correct options are A |z−2|=|z+2| C |z−1−i|=|z+1−i|Given : ω=(1−z)21−z2 ⇒ω=1−z1+z=z¯¯¯z−zz¯¯¯z+z⇒ω=¯¯¯z−1¯¯¯z+1=−¯¯¯ω⇒ω+¯¯¯ω=0 So, ω should be purely imaginary Hence, the locus of ω is x=0 Now, by obersvation, we see that all the options given represents perpendicular bisector. |z−2|=|z+2| Perpendicular bisector of points (2,0) and (−2,0) Which is x=0 |z−2i|=|z+2i| Perpendicular bisector of points (0,2) and (0,−2) Which is y=0 |z−1−i|=|z+1−i|⇒|z−(1+i)|=|z−(−1+i)| Perpendicular bisector of points (1,1) and (−1,1) Which is x=0 |z−i|=|z+i| Perpendicular bisector of points (0,1) and (0,−1) Which is y=0

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