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If $|z|=1$ and $z \neq \pm 1$, then all the points representing $\frac{z}{1-z^{2}}$ lie on;
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Verified Answer
The correct answer is:
the $y$-axis
Let $z=e^{i \theta}, \theta \neq \pm n \pi ~ n \in I$
Let $w=\frac{e^{i \theta}}{1-e^{i 2 \theta}}=\frac{1}{e^{-i \theta}-e^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta}$
$\therefore$ Locus is y-axis
Let $w=\frac{e^{i \theta}}{1-e^{i 2 \theta}}=\frac{1}{e^{-i \theta}-e^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta}$
$\therefore$ Locus is y-axis
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