Question

if $x_r$=cos$\left(\frac{\pi }{2^r}\right)$+isin$\left(\frac{\pi }{2^r}\right)$ , then $x_r$, $x_r$, ......$\infty$ is

• A. -3
• B. -2
• C. -1
• D. 0

Answer 1

The correct option is C

-1

$x_r$, $x_r$, ......Upto $\infty$ =$(cos\frac{\pi }{2}+isin\frac{\pi }{2})$$(cos\frac{\pi }{2^2}+isin\frac{\pi }{2^2})$

Upto........$\infty$

= cos$(\pi +\frac{\pi }{2}+\frac{\pi }{2^2}+.......)$ +isin$(\pi +\frac{\pi }{2}+\frac{\pi }{2^2}+.......)$

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