A) 18
B) 54
C) 6
D) 12
Correct Answer: D
Solution :
\[{{\operatorname{z}}^{2}} + z +1 = 0 \,\Rightarrow \,\,\,z = \omega or {{\omega }^{2}}\] So, \[\operatorname{z}+\frac{1}{z}=\omega +{{\omega }^{2}} =-1\] \[{{\operatorname{z}}^{2}}+\frac{1}{{{z}^{2}}}={{\omega }^{2}}+\omega =-1\] \[{{\operatorname{z}}^{3}}+\frac{1}{{{z}^{3}}}=\,\,{{\omega }^{3}}+{{\omega }^{3}} =\,\,2\] \[{{\operatorname{z}}^{4}}+\frac{1}{{{z}^{4}}}=\,-1,\,\,{{z}^{5}}+\frac{1}{{{z}^{2}}}=-1\,\,and\,\,{{z}^{6}}+\frac{1}{{{z}^{6}}}=2\] \[\therefore ~The given sum =1+1+4+1+1+4=12\]
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