AP EAMCET · Maths · Complex Number
If \(x_n=\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}\), then \(\prod_{n=1}^{\infty} x_n\) is equal to
- A \(-1\)
- B \(1\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\frac{i}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given that, }=\frac{9+6 i}{13} \\ & \qquad \begin{aligned} X_n & =\cos \frac{\pi}{2^n}+i \sin \frac{\pi}{2^n}=\operatorname{cis} \frac{\pi}{2^n} \\ \therefore \quad \prod_{n=1}^{\infty} X_n & =\prod_{n=1}^{\infty} \operatorname{cis} \frac{\pi}{2^n} \\ &…
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