If {z_r} = cos frac{{ralpha }}{{{n^2}}} + isi | vedclass

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(C) Given ${z_r} = \cos \frac{r\alpha}{n^2} + i\sin \frac{r\alpha}{n^2} = e^{i \frac{r\alpha}{n^2}}$.Then the product $P = {z_1}{z_2} \dots {z_n} = \p

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$e^{i\alpha / 2}$

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(C) Given ${z_r} = \cos \frac{r\alpha}{n^2} + i\sin \frac{r\alpha}{n^2} = e^{i \frac{r\alpha}{n^2}}$.Then the product $P = {z_1}{z_2} \dots {z_n} = \prod_{r=1}^{n} e^{i \frac{r\alpha}{n^2}} = e^{i \frac{\alpha}{n^2} \sum_{r=1}^{n} r}$.Using the sum formula $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$,we get:$P = e^{i \frac{\alpha}{n^2} \cdot \frac{n(n+1)}{2}} = e^{i \frac{\alpha(n^2+n)}{2n^2}} = e^{i \frac{\alpha}{2} (1 + \frac{1}{n})}$.Taking the limit as $n 	o \infty$:$\mathop {\lim }\limits_{n 	o \infty } P = e^{i \frac{\alpha}{2} (1 + 0)} = e^{i \frac{\alpha}{2}} = \cos \frac{\alpha}{2} + i\sin \frac{\alpha}{2}$.

If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}$,where $r = 1, 2, 3, \dots, n$,then $\mathop {\lim }\limits_{n \to \infty } {z_1}{z_2}{z_3} \dots {z_n}$ is equal to

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