• # question_answer If${{z}^{2}}+z+1=0,$where z is complex number, then the value of ${{\left( z+\frac{1}{z} \right)}^{2}}+{{\left( {{z}^{2}}+\frac{1}{{{z}^{2}}} \right)}^{2}}+{{\left( {{z}^{3}}+\frac{1}{{{z}^{3}}} \right)}^{2}}$ $+......+{{\left( {{z}^{6}}+\frac{1}{{{z}^{6}}} \right)}^{2}}$is       AIEEE  Solved  Paper-2006 A) 54                           B) 6                              C)        12           D)        18

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, ${{\left( z+\frac{1}{z} \right)}^{2}}+{{\left( {{z}^{2}}+\frac{1}{{{z}^{2}}} \right)}^{2}}+{{\left( {{z}^{3}}+\frac{1}{{{z}^{3}}} \right)}^{2}}$ $+{{\left( {{z}^{4}}+\frac{1}{{{z}^{4}}} \right)}^{2}}+{{\left( {{z}^{5}}+\frac{1}{{{z}^{5}}} \right)}^{2}}+{{\left( {{z}^{6}}+\frac{1}{{{z}^{6}}} \right)}^{2}}$ $={{(\omega +{{\omega }^{2}})}^{2}}+{{({{\omega }^{2}}+\omega )}^{2}}+{{({{\omega }^{3}}+{{\omega }^{-3}})}^{2}}$                 $+{{(\omega +{{\omega }^{2}})}^{2}}+{{({{\omega }^{2}}+\omega )}^{2}}+{{({{\omega }^{6}}+{{\omega }^{-6}})}^{2}}$ $={{(-1)}^{2}}+{{(-1)}^{2}}+{{(1+1)}^{2}}+{{(-1)}^{2}}+{{(-1)}^{2}}+{{(1+1)}^{2}}$ $=1+1+\text{ }4+1+1+4=12$