A) 0
B) 1
C) \[-1\]
D) \[i\]
Correct Answer: B
Solution :
We have, \[{{z}_{k}}=\cos \frac{\theta }{{{2}^{k}}}+i\,\sin \frac{\theta }{{{2}^{k}}}\] \[={{e}^{i\frac{\theta }{{{2}^{k}}}}}\] Put, \[k=1,2,3,....\infty \] \[{{z}_{1}}={{e}^{i\frac{\theta }{2}}},{{z}_{2}}={{e}^{i\frac{\theta }{{{2}^{2}}}}},{{z}_{3}}={{e}^{i\frac{\theta }{{{2}^{3}}}}}.,....\] \[\therefore \]\[{{z}_{1}}.{{z}_{2}}.{{z}_{3}}....={{e}^{\frac{i\theta }{2}}}.{{e}^{\frac{i\theta }{{{2}^{2}}}}}.{{e}^{\frac{i\theta }{{{2}^{3}}}}}....\] \[={{e}^{i\,\theta \,\left( \frac{1}{2}+\frac{1}{{{2}^{2}}}+\frac{1}{{{2}^{3}}}+... \right)}}\] \[={{e}^{i\,2n\pi \left( \frac{1/2}{1-1/2} \right)}}\] [\[\because \,\theta =2n\pi \]given] \[=\cos 2n\pi +i\sin 2n\pi \] \[=1\]
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