If Z_r = cos left(frac{pi}{2^r}right) + i sin | vedclass

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(D) Given $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right) = e^{i \frac{\pi}{2^r}}$.Therefore,the product $P = Z_1 Z_2 Z

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(D) Given $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right) = e^{i \frac{\pi}{2^r}}$.Therefore,the product $P = Z_1 Z_2 Z_3 \ldots \infty = e^{i \frac{\pi}{2^1}} \cdot e^{i \frac{\pi}{2^2}} \cdot e^{i \frac{\pi}{2^3}} \ldots$Using the property of exponents,$P = e^{i \left( \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \right)}$.The exponent is a geometric series with first term $a = \frac{\pi}{2}$ and common ratio $r = \frac{1}{2}$.The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{\pi/2}{1 - 1/2} = \frac{\pi/2}{1/2} = \pi$.Thus,$P = e^{i \pi} = \cos \pi + i \sin \pi$.Since $\cos \pi = -1$ and $\sin \pi = 0$,we get $P = -1 + i(0) = -1$.

If $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right)$ for $r = 1, 2, 3, \ldots$,then the product $Z_1 Z_2 Z_3 \ldots \infty$ is equal to:

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