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Question

If z=x+iy, |z|=1 and ω=(1z)21z2, then the locus of ω is equivalent to 
  1. |z2|=|z+2|
  2. |z2i|=|z+2i|
  3. |z1i|=|z+1i|
  4. |zi|=|z+i|


Solution

The correct options are
A |z2|=|z+2|
C |z1i|=|z+1i|
Given : ω=(1z)21z2
ω=1z1+z=z¯¯¯zzz¯¯¯z+zω=¯¯¯z1¯¯¯z+1=¯¯¯ωω+¯¯¯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.
|z2|=|z+2|
Perpendicular bisector of points (2,0) and (2,0)
Which is x=0

|z2i|=|z+2i|
Perpendicular bisector of points (0,2) and (0,2)
Which is y=0

|z1i|=|z+1i||z(1+i)|=|z(1+i)|
Perpendicular bisector of points (1,1) and (1,1)
Which is x=0

|zi|=|z+i|
Perpendicular bisector of points (0,1) and (0,1)
Which is y=0

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