A) \[\cos \alpha +i\,\sin \alpha \]
B) \[\cos (\alpha /2)-i\sin (\alpha /2)\]
C) \[{{e}^{i\alpha /2}}\]
D) \[\sqrt[3]{{{e}^{i\alpha }}}\]
Correct Answer: C
Solution :
\[{{z}_{r}}=\cos \frac{r\alpha }{{{n}^{2}}}+i\sin \frac{r\alpha }{{{n}^{2}}}\] \[{{z}_{1}}=\cos \frac{\alpha }{{{n}^{2}}}+i\sin \frac{\alpha }{{{n}^{2}}}\]; \[{{z}_{2}}=\cos \frac{2\alpha }{{{n}^{2}}}+i\sin \frac{2\alpha }{{{n}^{2}}}\]; .... Þ \[{{z}_{n}}=\,\cos \frac{n\alpha }{{{n}^{2}}}+i\sin \frac{n\alpha }{{{n}^{2}}}\] \[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\,({{z}_{1}}\,{{z}_{2}}\,{{z}_{3}}.........{{z}_{n}})\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \cos \,\left\{ \frac{\alpha }{{{n}^{2}}}(1+2+3+...+n) \right\} \right.\]\[\left. +i\sin \,\left\{ \frac{\alpha }{{{n}^{2}}}(1+2+3+...+n) \right\} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \cos \frac{\alpha \,n(n+1)}{2{{n}^{2}}}+i\sin \frac{\alpha \,n(n+1)}{2{{n}^{2}}} \right]\] \[=\,\cos \frac{\alpha }{2}+i\sin \frac{\alpha }{2}={{e}^{\frac{i\alpha }{2}}}\].
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