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WBJEE · Maths · Complex Number

If \(\theta \in \mathbb{R}\) and \(\frac{1-i \cos \theta}{1+2 i \cos \theta}\) is real number, then \(\theta\) will be (when \(I:\) Set of integers)

  1. A \((2 n+1) \frac{\pi}{2}, n \in l\)
  2. B \(\frac{3 n \pi}{2}, n \in I\)
  3. C \(n \pi, n \in I\)
  4. D \(2 n \pi, n \in l\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((2 n+1) \frac{\pi}{2}, n \in l\)

Step-by-step Solution

Detailed explanation

Let \(Z=\frac{1-i \cos \theta}{1+2 i \cos \theta}\) \(\therefore \quad \bar{z}=\frac{1+i \cos \theta}{1-2 i \cos \theta}\) Since, \(z\) is a real number, then \(z-\bar{z}=0\) \(\Rightarrow \frac{1-i \cos \theta}{1+2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta}\)…