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If omega=z/(z-1/3i) and abs(omega)=1, wh...

If `omega=z/(z-1/3i)` and `abs(omega)=1`, where `i=sqrt(-1)`,then lies on

Text Solution

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The correct Answer is:
Perpendicular bisector of the line joininig 0+ 0i and `0 + (1//3)i`
`omega = (z)/(z-(1)/(2)i)`
or ` |omega| = |(z)/(z-(1)/(3)i)|`
or ` |z| = |omega||z-(1)/(3)i|`
or ` |z| = |z-(1)/(3)i|" "(because |omega|=1)`
Hence, locous of z is perpendicular bisector of the line joining `0 + 0i` and `0+(1//3)i`. Hence z lies on a straight line .
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