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

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(C) Given ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}} = {e^{i\frac{{r\alpha }}{{{n^2}}}}}$.The product is ${z_1}{z_2}{

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${e^{i\alpha /2}}$

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(C) Given ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}} = {e^{i\frac{{r\alpha }}{{{n^2}}}}}$.The product is ${z_1}{z_2}{z_3}\dots{z_n} = \prod_{r=1}^{n} {e^{i\frac{{r\alpha }}{{{n^2}}}}} = {e^{i\frac{\alpha }{{{n^2}}}\sum_{r=1}^{n} r}}$.Using the sum formula $\sum_{r=1}^{n} r = \frac{{n(n + 1)}}{2}$,we get:${z_1}{z_2}{z_3}\dots{z_n} = {e^{i\frac{\alpha }{{{n^2}}} \cdot \frac{{n(n + 1)}}{2}}} = {e^{i\frac{{\alpha (n + 1)}}{{2n}}}} = {e^{i\frac{\alpha }{2}(1 + \frac{1}{n})}}$.Taking the limit as $n 	o \infty$:$\mathop {\lim }\limits_{n 	o \infty } {e^{i\frac{\alpha }{2}(1 + \frac{1}{n})}} = {e^{i\frac{\alpha }{2}(1 + 0)}} = {e^{i\frac{\alpha }{2}}} = \cos \frac{\alpha }{2} + i\sin \frac{\alpha }{2}$.

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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