If {x_n} = cos left( frac{pi }{4^n} right) + | vedclass

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(A) Given ${x_n} = \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right)$.Using the property of complex numbers,${x_1} \cdot {

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$\frac{1 + i\sqrt{3}}{2}$

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(A) Given ${x_n} = \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right)$.Using the property of complex numbers,${x_1} \cdot {x_2} \cdot {x_3} \dots \infty = \prod_{n=1}^{\infty} \left( \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right) \right)$.This is equal to $\cos \left( \sum_{n=1}^{\infty} \frac{\pi }{4^n} \right) + i \sin \left( \sum_{n=1}^{\infty} \frac{\pi }{4^n} \right)$.The sum of the infinite geometric series is $\sum_{n=1}^{\infty} \frac{\pi }{4^n} = \frac{\pi/4}{1 - 1/4} = \frac{\pi/4}{3/4} = \frac{\pi}{3}$.Therefore,the expression becomes $\cos \left( \frac{\pi}{3} \right) + i \sin \left( \frac{\pi}{3} \right)$.$= \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i\sqrt{3}}{2}$.

If ${x_n} = \cos \left( \frac{\pi }{4^n} \right) + i \sin \left( \frac{\pi }{4^n} \right)$,then ${x_1} \cdot {x_2} \cdot {x_3} \dots \infty = $

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