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If $\mid z\mid=1$ and $z\neq \pm 1$ then $\large\frac{z}{1-z^2}$ lie on

$\begin{array}{1 1}(A)x- axis \\ (B) y-axis \\ (C) \text{A line passing through origin} \\(D) |z|=2 \end{array}$

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$\omega=\large\frac{z}{1-z^2}=\frac{z}{z\overline{z}-z^2}=\frac{1}{\overline{z}-z}$
$\Rightarrow \omega=\large\frac{1}{-2iIm(z)}$
$\Rightarrow \omega$ is purely imaginary
$\Rightarrow \omega$ lies on $y$-axis
Hence (B) is the correct answer.
answered Apr 9, 2014 by sreemathi.v
 

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