Question

If $z_r=\cos \frac{r\alpha }{n^2}+i\sin \frac{r\alpha }{n^2}$, where $r=1,2,3,....n$, then $\underset{n\to \infty }{lim}\left(z_1.z_2.....z_n\right)$ is equal to

• A. $\cos \alpha +i\sin \alpha$
• B. $\sin \frac{\alpha }{2}+i\cos \frac{\alpha }{2}$
• C. $e^{i\alpha /2}$
• D. $\sqrt{e^{i\alpha }}$

Answer 1

The correct option is D $\sqrt{e^{i\alpha }}$

Solution -

$z_r=\frac{cosr\alpha }{n^2}+i\frac{sinr\alpha }{n^2}$

Comparing with $cos\theta +isin\theta$

$\theta =\frac{r\alpha }{n^2}$

$\therefore z_r=e^{\frac{ir\alpha }{n^2}}$

${lim}_{n\to \infty }(z_1,z_2,z_3....z_n)=e^{\frac{i\alpha }{n^2}}{e}^{\frac{2i\alpha }{n^2}}{e}^{\frac{3i\alpha }{n^2}}...e^{\frac{ni\alpha }{n^2}}$

$={lim}_{n\to \infty }{e}^{i\alpha (\frac{1}{n^2}+\frac{2}{n^2}+...)}$

$={lim}_{n\to \infty }{e}^{i\alpha \frac{1}{n^2}(1+2+3)}$

$={lim}_{n\to \infty }{e}^{i\alpha \left(\frac{n^2+n}{2n^{{}^2}}\right)}\dots S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

at $n\to \infty$

$=e^{i\alpha }\times \frac{1}{2}$

${lim}_{n\to \infty }(z_1,z_2,z_3....z_n)=\sqrt{e^{i\alpha }}$