A) 1
B) w
C) \[-1\]
D) 0
Correct Answer: B
Solution :
[b] Let \[y=\sqrt{-1-\sqrt{1-}\sqrt{-1}}..........+\infty \] \[y=\sqrt{-1-y}\] \[{{y}^{2}}=-1-y\] \[\Rightarrow {{y}^{2}}+y+1=0\] \[\therefore y=\frac{-1\pm \sqrt{1-4.1.1}}{2.1}\] \[=\frac{-1\pm \sqrt{3}}{2}=\frac{-1\pm \sqrt{3i}}{2}=\omega \,or{{\omega }^{2}}\] Let \[y=\frac{-1+\sqrt{3i}}{2}=\omega \] then \[y=\frac{-1-\sqrt{3}}{2}={{\omega }^{2}}\] Hence option [b] is correct.
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