Question

If $x_r=\cos \frac{\pi }{2^r}+i\sin \frac{\pi }{2^r}$, then $x_1.x_2.x_3....$ to $\infty =$

• A. 1/2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (B)

$x_r=\cos \frac{\pi }{2^r}+i\sin \frac{\pi }{2^r}$
$\Rightarrow x_{r=e^{i\pi /2^r}}$
Now . $x_1{x}_2{x}_3.....\infty =e^{i\pi /2}{e}^{i\pi /2^2}{e}^{i\pi /2^3}+.....$
$=e^{i\pi \left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....\right]}=e^{i\pi \left[\frac{\frac{1}{2}}{\frac{1-1}{2}}\right]}=e^{i\pi }$
$=\cos \pi +i\sin \pi =-1$