Question

If $z_n=\cos \frac{\pi }{(2n+1)(2n+3)}+i\sin \frac{\pi }{(2n+1)(2n+3)},n\in N,$ then the value of $\underset{k\to \infty }{lim}(z_1\cdot z_2\cdot z_3\cdots z_k)$ is

• A. $1$
• B. $\frac{\sqrt{3}}{2}+i\frac{1}{2}$
• C. $\frac{1}{2}+i\frac{\sqrt{3}}{2}$
• D. $\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}+i\frac{1}{2}$

$z_n=\cos \frac{\pi }{(2n+1)(2n+3)}+i\sin \frac{\pi }{(2n+1)(2n+3)}$
By using Euler's formula, we can write as
$z_n=e^{i{\theta }_n}$ where ${\theta }_n=\frac{\pi }{(2n+1)(2n+3)}$

Now,
$(z_1\cdot z_2\cdot z_3\cdots z_k)=e^{i\left(\sum _{n=1}^{n=k}{\theta }_n\right)}$
$=\cos \left(\sum _{n=1}^{n=k}\frac{\pi }{(2n+1)(2n+3)}\right)+i\sin \left(\sum _{n=1}^{n=k}\frac{\pi }{(2n+1)(2n+3)}\right)\cdots \left(i\right)$

Now,
$S_k=\sum _{n=1}^{n=k}\frac{\pi }{(2n+1)(2n+3)}$
$=\frac{\pi }{2}\sum _{n=1}^{n=k}[\frac{1}{2n+1}-\frac{1}{2n+3}]$
$=\frac{\pi }{2}[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+\cdots +(\frac{1}{2k+1}-\frac{1}{2k+3})]$
$=\frac{\pi }{2}[\frac{1}{3}-\frac{1}{2k+3}]$
$S_{\infty }=\underset{k\to \infty }{lim}\frac{\pi }{2}[\frac{1}{3}-\frac{1}{2k+3}]$
$=\frac{\pi }{6}$

From equation $\left(i\right)$,
$(z_1\cdot z_2\cdot z_3\cdots z_k)=\cos (\frac{\pi }{6}-\frac{\pi }{2(2k+3)})+i\sin (\frac{\pi }{6}-\frac{\pi }{2(2k+3)})$
$\Rightarrow \underset{k\to \infty }{lim}(z_1\cdot z_2\cdot z_3\cdots z_k)=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}=\frac{\sqrt{3}}{2}+i\frac{1}{2}$