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If |z|=1a n dz!=+-1, then all the values...

If `|z|=1a n dz!=+-1,` then all the values of `z/(1-z^2)` lie on a line not passing through the origin `|z|=sqrt(2)` the x-axis (d) the y-axis

A

a line not passing through the origin

B

`|z|=sqrt(2)`

C

the X-axies

D

the Y-axies

Text Solution

Verified by Experts

The correct Answer is:
D
Let `z=cos theta + I sin theta`
`rArr z/(1-z^2)=(cos theta + sin theta )/(1- (cos 2 theta + sin 2 theta))`
`=(cos theta + I sin theta )/(2 sin ^2 theta - 2 I sin theta cos theta)`
`=(cos theta +I sin theta)/(-2 I sin theta (cos theta +i sin theta))=i/(2 sin theta)`
Hence `z/1-z^2 ` lies on the imaginary axis ie, y-axis .
Alternate Solution
Let `E=(z)/(1-z^2)=z/(zbarz-z^2)=1/(barz -z)` which is an imaginary
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