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If $x_n=\cos \left(\frac{\pi}{4^n}\right)+i \sin \left(\frac{\pi}{4^n}\right)$, then $x_1 x_2 x_3 \ldots \infty$ is equal to
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Verified Answer
The correct answer is:
$\frac{1+i \sqrt{3}}{2}$
We have,
$$
\begin{aligned}
x_n & =e^i \frac{\pi}{4 n} \\
x_1 x_2 x_3 \ldots & =e^i \frac{\pi}{4} \cdot e^i \frac{\pi}{4^2} \cdot e^i \frac{\pi}{4^3} \ldots \\
& =e^{i \pi}\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3} \ldots\right) \\
=e^{i \pi} \frac{1 / 4}{1-\frac{1}{4}} & =e^i \frac{\pi}{3}=\cos 60^{\circ}+i \sin 60^{\circ} \\
=\frac{1}{2}+\frac{i \sqrt{3}}{2} & =\frac{1+i \sqrt{3}}{2}
\end{aligned}
$$
$$
\begin{aligned}
x_n & =e^i \frac{\pi}{4 n} \\
x_1 x_2 x_3 \ldots & =e^i \frac{\pi}{4} \cdot e^i \frac{\pi}{4^2} \cdot e^i \frac{\pi}{4^3} \ldots \\
& =e^{i \pi}\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3} \ldots\right) \\
=e^{i \pi} \frac{1 / 4}{1-\frac{1}{4}} & =e^i \frac{\pi}{3}=\cos 60^{\circ}+i \sin 60^{\circ} \\
=\frac{1}{2}+\frac{i \sqrt{3}}{2} & =\frac{1+i \sqrt{3}}{2}
\end{aligned}
$$
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