Question 390 of 455MediumMCQ
If z2+z+1=0, where z is complex number, then value of \[{\left( {z + {1 \over z}} \right)^2} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^2} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^2} + .......... + {\left( {{z^6} + {1 \over {{z^6}}}} \right)^2}\] is : Solutions · 1
Official solution
\[{z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,\] or \[\,\,\,{\omega ^2}\]
So, \[z + {1 \over z} = \omega + {\omega ^2} = - 1\]
\[{z^2} + {1 \over {{z^2}}} = {\omega ^2} + \omega = - 1,\]
\[{z^3} + {1 \over {{z^3}}} = {\omega ^3} + {\omega ^3} = 2\]
\[{z^4} + {1 \over {{z^4}}} = - 1,\]
\[{z^5} + {1 \over {{z^5}}} = - 1\]
and \[\,\,\,\,{z^6} + {1 \over {{z^6}}} = 2\]
\[\therefore\] The given sum =1+1+4+1+1+4=12 Answer the question to unlock