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JEE Advanced · Mathematics · 3. Complex Numbers

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

  1. A a line not passing through the origin
  2. B \(|z|=\sqrt{2}\)
  3. C the X-axis
  4. D the Y-axis
Verified Solution

Answer & Solution

Correct Answer

(D) the Y-axis

Step-by-step Solution

Detailed explanation

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.
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