Question

If $x_r=\cos \left(\frac{\pi}{2^r}\right)+i \sin \left(\frac{\pi}{2^r}\right)$, then $x_1 \cdot x_2 \ldots \ldots \infty$ is

(A) -3

(B) -2

(C) -1

(D) 0

Answer 1

$\therefore x_1 \cdot x_2 \cdot x_3 \ldots \ldots \infty$

$=\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\left(\cos \frac{\pi}{2^2}+i \sin \frac{\pi}{2^2}\right) \ldots \infty$

$=\cos \left(\frac{\pi}{2}+\frac{\pi}{2^2}+\ldots \ldots\right)+i \sin \left(\frac{\pi}{2}+\frac{\pi}{2^2}+\ldots\right)$

$=\cos \left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)+i \sin \left(\frac{\frac{\pi}{2}}{1-\frac{1}{2}}\right)$

$=\cos \pi+i \sin \pi=-1$

So, The correct option will be (C).