A) \[-1\]
B) \[1\]
C) \[0\]
D) \[\sqrt{2}\]
Correct Answer: A
Solution :
\[\therefore \]\[\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\left( \cos \left( \frac{\pi }{{{2}^{2}}} \right)+i\sin \left( \frac{\pi }{{{2}^{2}}} \right) \right)\] \[\left( \cos \left( \frac{\pi }{{{2}^{3}}} \right)+i\sin \left( \frac{\pi }{{{2}^{3}}} \right) \right)...\infty \] \[=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+...\infty \right)\] \[=\cos \left( \frac{{{\frac{\pi }{2}}^{1}}}{1-\frac{1}{2}} \right)+i\sin \left( \frac{{{\frac{\pi }{2}}^{1}}}{1-\frac{1}{2}} \right)\] \[=\cos \pi +i\sin \pi \] \[=-1\] Alternative Solution: \[\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\left( \cos \left( \frac{\pi }{{{2}^{2}}} \right)+i\sin \left( \frac{\pi }{{{2}^{2}}} \right) \right)\] \[\left( \cos \left( \frac{\pi }{{{2}^{3}}} \right)+i\sin \left( \frac{\pi }{{{2}^{3}}} \right) \right)...\infty \] \[={{e}^{\frac{i\pi }{e}}}\cdot {{e}^{\frac{i\pi }{{{2}^{2}}}}}\cdot {{e}^{\frac{i\pi }{{{2}^{3}}}}}...\infty \] \[={{e}^{\frac{i\pi }{2}}}\left( 1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+...\infty \right)\] \[={{e}^{\frac{i\pi }{2}\left( \frac{1}{1-\frac{1}{2}} \right)}}={{e}^{i\pi }}\] Note: De-moivre?s is used only when real part is cosine and imaginary part is sine form with same angle.
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