For n in N,if A_n = cos left(frac{pi}{2^n}rig | vedclass

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(D) Given that $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right) = e^{i \frac{\pi}{2^n}}$.Then,$A_1 A_2 A_3 A_4 = e^{i \p

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$\frac{1-i}{\sqrt{2}}$

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(D) Given that $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right) = e^{i \frac{\pi}{2^n}}$.Then,$A_1 A_2 A_3 A_4 = e^{i \pi \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right)}$.Summing the exponents: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{8+4+2+1}{16} = \frac{15}{16}$.So,$(A_1 A_2 A_3 A_4)^4 = (e^{i \pi \frac{15}{16}})^4 = e^{i \pi \frac{15}{4}}$.$e^{i \frac{15\pi}{4}} = e^{i (4\pi - \frac{\pi}{4})} = \cos(4\pi - \frac{\pi}{4}) + i \sin(4\pi - \frac{\pi}{4})$.$= \cos(\frac{\pi}{4}) - i \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} = \frac{1-i}{\sqrt{2}}$.

For $n \in N$,if $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right)$,then $(A_1 A_2 A_3 A_4)^4 =$

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