If x_n = cos frac{pi}{2^n} + i sin frac{pi}{2 | vedclass

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(A) Given that $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.We need to find the product $P = \prod_{n=1}^{\infty} x_n

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$-1$

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(A) Given that $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.We need to find the product $P = \prod_{n=1}^{\infty} x_n = \prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}}$.Using the property of exponents,$P = e^{i \sum_{n=1}^{\infty} \frac{\pi}{2^n}}$.The sum in the exponent is a geometric series: $\sum_{n=1}^{\infty} \frac{\pi}{2^n} = \pi \left( \frac{1/2}{1 - 1/2} \right) = \pi \left( \frac{1/2}{1/2} \right) = \pi$.Therefore,$P = e^{i \pi}$.Using Euler's formula,$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.

If $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n}$,then $\prod_{n=1}^{\infty} x_n$ is equal to

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