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