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) $|z|=\sqrt{2}$
(2007, 3M)
(c) the $X$-axis
(d) the $Y$-axis
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Answer:
Correct Answer: 11. (d)
Solution:
- 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.
