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Question

If $$|\mathrm{z}|=1$$ and $$\mathrm{z}\neq\pm 1$$, then all the values of $$\displaystyle \frac{\mathrm{z}}{1-\mathrm{z}^{2}}$$ lie on 


A
a line not passing through the origin
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B
|z|=2
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C
the x-axis
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D
the yaxis
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Solution

The correct option is D the $$y-$$axis
Let $$\displaystyle \mathrm{z}=\cos\theta+i\sin\theta$$, 
Hence, 
 $$\displaystyle\frac{\mathrm{z}}{1-\mathrm{z}^{2}}$$
$$=\dfrac{\cos\theta+i \sin\theta}{1-(\cos 2\theta+\mathrm{i}\sin 2\theta)}$$
$$\displaystyle =\frac{\cos\theta+\mathrm{i}\sin\theta}{2\sin^{2}\theta-2\mathrm{i}\sin\theta\cos\theta}$$
$$=\dfrac{\cos\theta+\mathrm{i}\sin\theta}{-2\mathrm{i}\sin\theta(\cos\theta+\mathrm{i}\sin\theta)}$$
$$\displaystyle =\frac{\mathrm{i}}{2\sin\theta}$$
Hence, $$\displaystyle \frac{\mathrm{z}}{1-\mathrm{z}^{2}}$$ lies on the imaginary axis i.e., $$\mathrm{y}=0$$.

Mathematics

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