A) \[1\]
B) \[-i\]
C) \[i\]
D) \[-1\]
Correct Answer: D
Solution :
Given\[{{z}_{r}}=\cos \frac{\pi }{{{2}^{r}}}+i\sin \frac{\pi }{{{2}^{r}}};\,\,r=1,\,\,2,\,\,...\] \[\therefore \]\[{{z}_{1}}{{z}_{2}}{{z}_{3}}...=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right)...\]\[={{e}^{i\pi /2\left( 1+\frac{1}{2}+\frac{1}{4}+......\infty \right)}}={{e}^{i\pi /2\frac{1}{1-1/2}}}={{e}^{i\pi /2\cdot 2}}={{e}^{i\pi }}\] \[=\cos \pi +i\sin \pi =-1\] Alternate Method: \[{{z}_{1}}{{z}_{2}}.....{{z}_{n}}=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+.... \right)\] \[+i\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+.... \right)\] ( By Applying De-Moivre's theorem) \[\cos \left( \frac{\pi }{2}\left( 1+\frac{1}{2}+\frac{1}{2}+.......\infty \right) \right)\] \[+i\sin \left( \frac{\pi }{2}\left( 1+\frac{1}{2}+.......\infty \right) \right)\] \[=\cos \left( \frac{\pi }{2}\cdot \frac{1}{1-\frac{1}{2}} \right)+i\sin \left( \frac{\pi }{2}\cdot \frac{1}{1-\frac{1}{2}} \right)\] \[=\cos \pi +i\sin \pi =-1\]
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