A) 54
B) 6
C) 12
D) 18
Correct Answer: C
Solution :
Given equation is \[{{z}^{2}}+z+1=0\] \[z=\frac{-1\pm \sqrt{1-4\times 1\times 1}}{2\times 1}\] \[z=\frac{-1\pm \sqrt{-3}}{2}\] \[\Rightarrow \]\[z=\omega ,{{\omega }^{2}}\] Now, \[{{\left( z+\frac{1}{z} \right)}^{2}}+{{\left( {{z}^{2}}+\frac{1}{{{z}^{2}}} \right)}^{2}}+{{\left( {{z}^{3}}+\frac{1}{{{z}^{3}}} \right)}^{2}}\] \[+{{\left( {{z}^{4}}+\frac{1}{{{z}^{4}}} \right)}^{2}}+{{\left( {{z}^{5}}+\frac{1}{{{z}^{5}}} \right)}^{2}}+{{\left( {{z}^{6}}+\frac{1}{{{z}^{6}}} \right)}^{2}}\] \[={{(\omega +{{\omega }^{2}})}^{2}}+{{({{\omega }^{2}}+\omega )}^{2}}+{{({{\omega }^{3}}+{{\omega }^{-3}})}^{2}}\] \[+{{(\omega +{{\omega }^{2}})}^{2}}+{{({{\omega }^{2}}+\omega )}^{2}}+{{({{\omega }^{6}}+{{\omega }^{-6}})}^{2}}\] \[={{(-1)}^{2}}+{{(-1)}^{2}}+{{(1+1)}^{2}}+{{(-1)}^{2}}+{{(-1)}^{2}}+{{(1+1)}^{2}}\] \[=1+1+\text{ }4+1+1+4=12\]
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