TS EAMCET · Maths · Complex Number
For \(n \in \mathbf{N}\), If \(A_n=\cos \left(\frac{\pi}{2^n}\right)+i \sin \left(\frac{\pi}{2^n}\right)\), then \(\left(A_1 A_2 A_3 A_4\right)^4=\)
- A \(\frac{-1-i}{\sqrt{2}}\)
- B 1
- C 0
- D \(\frac{1-i}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1-i}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
It is given that, \(A_n=\cos \left(\frac{\pi}{2^n}\right)+i \sin \left(\frac{\pi}{2^n}\right), n \in \mathbf{N}=e^{i\left(\frac{\pi}{2^n}\right)}\) \(\therefore\left(A_1 A_2 A_3 A_4\right)^4=e^{i \pi\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)^4}\)…
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