WBJEE · Maths · Complex Number
If \(|z|=1\) and \(z \neq \pm 1\), then all the points representing \(\frac{z}{1-z^{2}}\) lie on;
- A a line not passing through the origin
- B the line \(y=x\)
- C the \(x\)-axis
- D the \(y\)-axis
Answer & Solution
Correct Answer
(D) the \(y\)-axis
Step-by-step Solution
Detailed explanation
Let \(z=e^{i \theta}, \theta \neq \pm n \pi ~ n \in I\) Let \(w=\frac{e^{i \theta}}{1-e^{i 2 \theta}}=\frac{1}{e^{-i \theta}-e^{i \theta}}=\frac{1}{-2 \cdot i \sin \theta}\) \(\therefore\) Locus is y-axis
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