Complex Numbers And Quadratic Equations question 206
Question: If $ x_{r}=\cos ( \frac{\pi }{2^{r}} )+i\sin ( \frac{\pi }{2^{r}} ) $ , then $ x_1.x_2……\infty $ is [RPET 1990, 2000; BIT Mesra 1996; Karnataka CET 2000]
Options:
A) $ -3 $
B) $ -2 $
C) $ -1 $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
$ x_1,x_2,x_3….. $ upto $ \infty = $ $ ( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} )( \cos \frac{\pi }{2^{2}}+i\sin \frac{\pi }{2^{2}} ) $ upto ….. $ \infty $ $ =\cos ( \frac{\pi }{2}+\frac{\pi }{2^{2}}+….. ) $ $ +i\sin ( \frac{\pi }{2}+\frac{\pi }{2^{2}}+….. ) $ $ =\cos ( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} )+i\sin ( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} ) $ $ =\cos \pi +i\sin \pi =-1 $ .
BETA
