A) zero
B) 1
C) \[\pi \]
D) \[-1\]
Correct Answer: D
Solution :
\[{{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right)\] \[{{r}_{1}}=\cos \left( \,\frac{\pi }{2} \right)+i\sin \left( \frac{\pi }{2} \right)\] \[{{x}_{2}}=\cos \left( \frac{\pi }{{{2}^{2}}} \right)+i\sin \left( \frac{\pi }{{{2}^{2}}} \right),\] \[{{x}_{3}}=\cos \left( \frac{\pi }{{{2}^{3}}} \right)+i\sin \left( \frac{\pi }{{{2}^{3}}} \right)\] \[{{x}_{1}},{{x}_{2}},{{x}_{3}}...\infty \] \[=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+... \right)+i\] \[\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}} \right)\] \[=\cos \left( \frac{\pi /2}{1-1/2} \right)+i\sin \left( \frac{\pi /2}{1-1/2} \right)\] \[=\cos \pi +i\sin \pi =-1\]
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