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
B
|z|=2
C
the x-axis
D
the yaxis

Solution

## The correct option is D the $$y-$$axisLet $$\displaystyle \mathrm{z}=\cos\theta+i\sin\theta$$, 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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