Question

If $\left|z\right|=1$ and $z\ne \pm 1$, then all the values of $\frac{z}{1-z^2}$ lie on

• A. a line not passing through the origin
• B. $\left|z\right|=\sqrt{2}$
• C. the $x-$ axis
• D. the $y-$ axis

Answer 1

Answer: (D)

the $y-$ axis

Ans. $\left(d\right)$.
Since $\left|z\right|=1$ $\therefore$ $z=e^{i\theta }$ as $r=1$
$\therefore$ $\frac{z}{1-z^2}=\frac{1}{z^{-1}-z}=\frac{1}{e^{-i\theta }-e^{i\theta }}=\frac{-1}{2i\sin \theta }$
$=0+i$, $\frac{1}{2\sin \theta }\dots \frac{1}{i}=-i$
$\therefore$ Real part of $\frac{z}{1-z^2}$ is zero, Hence it lies on $y-$axis.