MHT CET · Maths · Complex Number
\(\mathrm{z}=\frac{3+2 i \sin \theta}{1-2 i \sin \theta}, \quad(i=\sqrt{-1})\) will be purely imaginary if \(\theta=\)
- A \(2 n \pi \pm \frac{\pi}{8}\), where \(n \in \mathbb{Z}\)
- B \(\mathrm{n} \pi+\frac{\pi}{8}\), where \(\mathrm{n} \in \mathbb{Z}\)
- C \(n \pi \pm \frac{\pi}{3}\), where \(n \in \mathbb{Z}\)
- D \(n \pi\), where \(n \in \mathbb{Z}\)
Answer & Solution
Correct Answer
(C) \(n \pi \pm \frac{\pi}{3}\), where \(n \in \mathbb{Z}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{z}=\frac{3+2 i \sin \theta}{1-2 i \sin \theta} \cdot \frac{1+2 i \sin \theta}{1+2 i \sin \theta}\) \(\mathrm{z}=\frac{3+6 i \sin \theta+2 i \sin \theta-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}\)
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