Question

If $z_r=CiS\left(\frac{\pi }{2^r}\right)$ for $1\le r\le n;r,n\in N$ then -

• A. $\underset{n\to \infty }{lim}\left(\prod _{r=1}^n{z}_r\right)=-1$
• B. $\underset{n\to \infty }{lim}\left(\prod _{r=1}^n{z}_r\right)=0$
• C. $\underset{n\to \infty }{lim}\left(\prod _{r=1}^n{z}_r\right)=1$
• D. $\underset{n\to \infty }{lim}\left(\prod _{r=1}^n{z}_r\right)=2$

Answer 1

The correct option is A $\underset{n\to \infty }{lim}\left(\prod _{r=1}^n{z}_r\right)=-1$

$z_r=Cis\left(\frac{\pi }{2^r}\right)$

$z_1{z}_2{z}_3...\infty =\cos (\frac{\pi }{2}+\frac{\pi }{2^2}+\frac{\pi }{2^3}+...\infty )+i\sin (\frac{\pi }{2}+\frac{\pi }{2^2}+\frac{\pi }{2^3}+...\infty )$

$=\cos \frac{\frac{\pi }{2}}{1-\frac{1}{2}}+i\sin \frac{\frac{\pi }{2}}{1-\frac{1}{2}}$

$=\cos (\frac{\pi }{2}\times 2)+i\sin (\frac{\pi }{2}\times 2)$

$=\cos \pi +i\sin \pi =-1+i\times 0=-1$

${lim}_{n\to \infty }{\prod }_{r=1}^n{z}_r=-1$