Question

If $z_r=\cos \left(\frac{\pi }{2^r}\right)+i\sin \left(\frac{\pi }{2^r}\right),$ then $z_1.z_2.z_3$ upto $\infty$ equals:

• A. $-3$
• B. $-2$
• C. 1
• D. 0
• E. -1

Answer 1

Answer: (E)

$z_1.z_2.z_3....\infty$
$=\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\left(\cos \frac{\pi }{2^2}+i\sin \frac{\pi }{2^2}\right)$ $\times \left(\cos \frac{\pi }{2^3}+i\sin \frac{\pi }{2^3}\right)...$
$=\cos \left(\frac{\pi }{2}+\frac{\pi }{2^2}+\frac{\pi }{2^3}+....\right)$ $+i\sin \left(\frac{\pi }{2}+\frac{\pi }{2^2}+\frac{\pi }{2^3}+....\right)$ $=\cos \left(\frac{\pi /2}{1-1/2}\right)+i\sin \left(\frac{\pi /2}{1-1/2}\right)$ $=\cos \pi +i\sin \pi =-1$