Question

If \(|z|=1\) and \(z \neq \pm 1\), then all the points representing \(\frac{z}{1-z^{2}}\) lie on;
(A) a line not passing through the origin
(B) the line \(y=x\)
(C) the \(x\)-axis
(D) the \(y\)-axis

Answer 1

Ans: (D)
Hint : Let \(z=c^{i \theta}, \theta \pm h \pi\)
$$
\begin{aligned}
&\text { Let } w=\frac{c^{i \theta}}{1-c^{i z \theta}}=\frac{1}{c^{-i \theta}-c^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta} \\
&\Rightarrow w=\frac{i}{2 \sin \theta} \\
&\longrightarrow \text { Purely imaginary } \\
&\therefore \text { Locus is } y \text {-axis }
\end{aligned}
$$