Question

If $z_r=cos\frac{r\alpha }{n^2}+isin\frac{r\alpha }{n^2}$, where $r=1$, $2$, $3$, $....$, $n$, then $\underset{n\to \infty }{\lim }{z}_1{z}_2{z}_3...z_n$ is equal to

• A. $cos\alpha +isin\alpha$
• B. $cos\left(\alpha /2\right)-isin\left(\alpha /2\right)$
• C. $e^{i\alpha /2}$
• D. $\sqrt[3]{e^{i\alpha }}$

Answer 1

Answer: (C)

$z_r=cos\frac{r\alpha }{n^2}+isin\frac{r\alpha }{n^2}$
$z_1=cos\frac{\alpha }{n^2}+isin\frac{\alpha }{n^2};$
$z_2=cos\frac{2\alpha }{n^2}+isin\frac{2\alpha }{n^2};...$
$\Rightarrow z_n=cos\frac{n\alpha }{n^2}+isin\frac{n\alpha }{n^2}$
consider $\underset{n\to \infty }{\lim }\left(z_1{z}_2{z}_3....z_n\right)$
$=\underset{n\to \infty }{\lim }\left[cos\left\{\frac{\alpha }{n^2}\left(1+2+3+....+n\right)\right\}+isin\left\{\frac{\alpha }{n^2}\left(1+2+3+....+n\right)\right\}\right]$
$=\underset{n\to \infty }{\lim }\left[cos\frac{\alpha n\left(n+1\right)}{2n^2}+isin\frac{\alpha n\left(n+1\right)}{2n^2}\right]$
$=\underset{n\to \infty }{\lim }\frac{cos\alpha \left(1+\frac{1}{n}\right)}{2}+\frac{isin\alpha \left(1+\frac{1}{n}\right)}{2}$
$=cos\frac{\alpha }{2}+isin\frac{\alpha }{2}=e^{\frac{i\alpha }{2}}$