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$