A) \[\frac{\pi }{6}\,\]
B) \[\frac{\pi }{3}\,\]
C) 0
D) \[\frac{\pi }{4}\,\]
Correct Answer: D
Solution :
\[1+x+{{x}^{2}}=0\] \[x\,\,=\,\,\frac{-1\pm \,\sqrt{1-4}}{2}\,\,=\,\frac{-1\,\pm \,i\sqrt{3}}{2}\] \[{{z}_{0}}=w,\text{ }{{w}^{2}}\] Now \[z=3+6i\,{{z}_{\text{0}}}^{\text{81}}-3i\,{{z}_{0}}^{93}\] \[z=3+6i{{w}_{{}}}^{81}-3i{{w}^{9}}^{3}\left( {{w}^{93}}={{w}^{8}}^{1}=1 \right)\] \[\Rightarrow \,\,\,\,\,\,z=3+3i\] then arg(z) \[~=ta{{n}^{-1}}\left( \frac{3}{3} \right)=ta{{n}^{-1}}\left( 1 \right)=\frac{\pi }{4}\]
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