JEE MAIN - Mathematics (2006 - No. 16)
If $${z^2} + z + 1 = 0$$, 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 :
18
54
6
12
설명
$${z^2} + z + 1 = 0 \Rightarrow z = \omega \,\,\,$$ 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 $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$
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 $$ = 1 + 1 + 4 + 1 + 1 + 4 = 12$$