Question
If $$\left| z \right| = 1\,{\text{and }}z \ne \pm 1,$$ 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 :
the $$y$$ - axis
Solution :
$${\text{Given }}\left| z \right| = 1\,{\text{and }}z \ne \pm 1$$
To find locus of $$\omega = \frac{z}{{1 - {z^2}}}$$
We have $$\omega = \frac{z}{{1 - {z^2}}} = \frac{z}{{z\overline z - {z^2}}}\,\,\,\,\,\,\,\,\left[ {\because \,\left| z \right| = 1\,\, \Rightarrow {{\left| z \right|}^2} = z\overline z = 1} \right]$$
$$\,\,\,\,\, = \frac{1}{{\overline z - z}}$$
= purely imaginary number
∴ $$\omega $$ must lie on $$y$$ - axis.