• # question_answer If ${{z}_{r}}=\cos \frac{r\alpha }{{{n}^{2}}}+i\sin \frac{r\alpha }{{{n}^{2}}},$ where r = 1, 2, 3,?.,n, then $\underset{n\to \infty }{\mathop{\lim }}\,\,\,{{z}_{1}}{{z}_{2}}{{z}_{3}}...{{z}_{n}}$ is equal to [UPSEAT 2001] A) $\cos \alpha +i\,\sin \alpha$ B) $\cos (\alpha /2)-i\sin (\alpha /2)$ C) ${{e}^{i\alpha /2}}$ D) $\sqrt[3]{{{e}^{i\alpha }}}$

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