Question

If $|z|=1$ 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. $|z|=\sqrt{2}$
• C. the $x$-axis
• D. the $y-$axis

Answer 1

The correct option is D the $y-$axis

Let $z=\cos \theta +i\sin \theta$,

Hence,

$\frac{z}{1-z^2}$

$=\frac{\cos \theta +i\sin \theta }{1-(\cos 2\theta +i\sin 2\theta )}$
$=\frac{\cos \theta +i\sin \theta }{2{\sin }^2\theta -2i\sin \theta \cos \theta }$

$=\frac{\cos \theta +i\sin \theta }{-2i\sin \theta (\cos \theta +i\sin \theta )}$
$=\frac{i}{2\sin \theta }$
Hence, $\frac{z}{1-z^2}$ lies on the imaginary axis i.e., $y=0$.