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If z!=1 and (z^2)/(z-1) is real, then th...

If `z!=1` and `(z^2)/(z-1)` is real, then the point represented by the complex number z lies (1) either on the real axis or on a circle passing through the origin (2) on a circle with centre at the origin (3) either on the real axis or on a circle not passing through the origin (4) on the imaginary axis

A

either on the real axis or on a circle passing thorugh the origin.

B

on a circle with centre at the origin.

C

either on the real axis or an a circle not possing through the origin .

D

on the imaginary axis .

Text Solution

Verified by Experts

The correct Answer is:
A

`(z^(2))/(z-1) ` is purely real lt
`therefore (z^(2))/(z-1) = (barz^(2))/(barz-1)`
` rArr zbarzz - z^(2) = zbarzbar - barz^(-2)`
`rArr (z-barz) (|z|^(2) -( z+ barz))= 0`
Either `z = barz`
`rArr ` z lies on real axis.
or `|z|^(2) = z+z`
`rArr zbarz - z - barz = 0`
`rArr x^(2) + y^(2) -2x = 0`
Which represents a cricle passing throught origin.
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