Question

$x_n=\cos \frac{\pi }{2^n}+i\sin \frac{\pi }{2^n}$ then $x_1{x}_2{x}_3{x}_4......x_{\infty }=$

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

Answer: (B)

$-1$

$x_n=\cos \frac{\pi }{2^n}+i\sin \frac{\pi }{2^n}$

$x_1{x}_2{x}_3{x}_4......x_{\infty }=\text{cis}\frac{\pi }{2}\cdot \text{cis}\frac{\pi }{2^2}\cdot \text{cis}\frac{\pi }{2^3}.....\infty$

$=\text{cis}(\frac{\pi }{2}+\frac{\pi }{2^2}+\frac{\pi }{2^3}.....)$

$=\text{cis}\left(\frac{\pi }{2}\right(1+\frac{1}{2}+\frac{1}{2^2}+.....\left)\right)$

$\Rightarrow \text{cis}\pi =-1$

Hence, option B.