Complex Numbers 2 Question 11

11.

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

(d) the $Y$-axis

(2007, 3M)

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Answer:

Correct Answer: 11. (d)

Solution:

  1. Let $z=\cos \theta+i \sin \theta$

$ \begin{aligned} \quad \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. $Y$-axis.

Alternate Solution

Let $E=\frac{z}{1-z^{2}}=\frac{z}{z \bar{z}-z^{2}}=\frac{1}{\bar{z}-z}$ which is an imaginary.



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