A) \[-1\]
B) \[0\]
C) \[-\,\,i\]
D) \[i\]
Correct Answer: D
Solution :
\[{{z}_{r}}=\cos \,\left( \frac{\pi }{{{3}^{r}}} \right)+i\,\,\sin \,\left( \frac{\pi }{{{3}^{r}}} \right)\] \[{{z}_{r}}={{e}^{i\,\pi /{{3}^{r}}}},\,\,{{z}_{1}}={{e}^{2/3}},\,{{z}_{2}}={{e}^{i\,z/{{3}^{2}}}},{{z}_{3}}={{e}^{i\,z/{{3}^{3}}}},.....\] so on. Now, \[{{z}_{1}}.{{z}_{2}}.{{z}_{3}}...\infty ={{e}^{i\pi /3}}.{{e}^{i\pi /{{3}^{2}}}}.{{e}^{i\pi /{{3}^{3}}}}.....\infty \] \[={{e}^{i\pi \,(1/3+1/{{3}^{2}}+1/{{3}^{3}}.+....\infty )}}\] \[=e{{\,}^{i\,\pi }}\,\left\{ \frac{1/3}{1-1/3} \right\}\] \[\Rightarrow \] \[={{e}^{i\pi \times 1/2}}=i\]
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