Question

If $z_{r} = cos \frac{r\alpha}{n^{2}}+i\,sin \frac{r\alpha }{n^{2}}$, where $r = 1$, $2$, $3$, $....$, $n$, then $\displaystyle\lim_{n\to\infty} z_{1}z_{2}z_{3} ... z_{n}$ is equal to

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

Answer 1

Answer: (C)

$z_{r} = cos \frac{r\alpha}{n^{2}}+ i\,sin \frac{r\alpha }{n^{2}}$
$z_{1} = cos\frac{\alpha }{n^{2}} + i\,sin\frac{\alpha }{n^{2}} ; $
$z_{2} = cos \frac{2\alpha }{n^{2}}+i\,sin\frac{2\alpha }{n^{2}}; ...$
$\Rightarrow z_{n} = cos \frac{n\alpha }{n^{2}} + i\,sin \frac{n\alpha }{n^{2}}$
consider $\displaystyle\lim_{n\to\infty}\left(z_{1} \,z_{2} \,z_{3} .... z_{n} \right)$
$= \displaystyle\lim _{n\to \infty }\left[ cos \left\{\frac{\alpha }{n^{2}}\left(1+2+3+ .... +n\right)\right\}+i\,sin\left\{\frac{\alpha }{n^{2}}\left(1+2+3+ .... +n\right)\right\}\right]$
$= \displaystyle\lim _{n\to \infty } \left[cos \frac{\alpha n\left(n+1\right)}{2n^{2}}+i\,sin\frac{\alpha n\left(n+1\right)}{2n^{2}}\right]$
$= \displaystyle\lim _{n\to \infty } \frac{cos\alpha\left(1+\frac{1}{n}\right)}{2} + \frac{i\,sin\alpha\left(1+\frac{1}{n}\right)}{2}$
$= cos \frac{\alpha}{2} + i\,sin \frac{\alpha}{2} = e^{\frac{i\alpha}{2}}$