If {x_r} = cos(pi/3^r) - isin(pi/3^r) (where | vedclass

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(C) Given ${x_r} = \cos(\pi/3^r) - i\sin(\pi/3^r)$.We need to find the product $P = x_1 \cdot x_2 \cdot x_3 \cdots \infty$.Using the property of compl

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$-i$

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(C) Given ${x_r} = \cos(\pi/3^r) - i\sin(\pi/3^r)$.We need to find the product $P = x_1 \cdot x_2 \cdot x_3 \cdots \infty$.Using the property of complex numbers,$(\cos 	heta_1 - i\sin 	heta_1)(\cos 	heta_2 - i\sin 	heta_2) = \cos(	heta_1 + 	heta_2) - i\sin(	heta_1 + 	heta_2)$.Thus,$P = \cos(\sum_{r=1}^{\infty} \frac{\pi}{3^r}) - i\sin(\sum_{r=1}^{\infty} \frac{\pi}{3^r})$.The sum in the argument is an infinite geometric series: $S = \frac{\pi}{3} + \frac{\pi}{9} + \frac{\pi}{27} + \cdots$.Here,the first term $a = \pi/3$ and the common ratio $r = 1/3$.The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{\pi/3}{1 - 1/3} = \frac{\pi/3}{2/3} = \frac{\pi}{2}$.Therefore,$P = \cos(\pi/2) - i\sin(\pi/2) = 0 - i(1) = -i$.

If ${x_r} = \cos(\pi/3^r) - i\sin(\pi/3^r)$ (where $i = \sqrt{-1}$),then the value of $x_1 \cdot x_2 \cdot x_3 \cdots \infty$ is:

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