Q27 / 66 · JEE Advanced 2007 (Paper 2) · Mathematics

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

  • A
    a line not passing through the origin
  • B
    $|z|=\sqrt{2}$
  • C
    the X-axis
  • Correct answer
    D
    the Y-axis

Solution

Let $z=\cos \theta+i \sin \theta$ $$ \begin{aligned} \frac{z}{1-z^2} & =\frac{\cos \theta+i \sin \theta}{1-(\cos 2 \theta+i \sin 2 \theta)} \\ & =\frac{\cos \theta+i \sin \theta}{2 \sin ^2 \theta-2 i \sin \theta \cos \theta} \\ & =\frac{\cos \theta+i \sin \theta}{-2 i \sin \theta(\cos \theta+i \sin \theta)} \\ & =\frac{i}{2 \sin \theta} \end{aligned} $$ Hence, $\frac{z}{1-z^2}$ lies on the imaginary axis, i.e. $x=0$. ALITER Let $E=\frac{z}{1-z^2}=\frac{z}{z \bar{z}-z^2}=\frac{1}{\bar{z}-z}$ which is imaginary.

Hence, the correct answer is option: 'D'.

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