If z2+z+1=0,where z is complex number,then find value of (z+1/z)² + (z²+1/z²)² + (z³+1/z³)² + ....... + (z^6+1/z^6)²?

1 Answer
Shwetank Mauria
Mar 27, 2018

#(z+1/z)^2+(z^2+1/z^2)^2+(z^3+1/z^3)^2+.....+(z^6+1/z^6)^2=12#

Explanation:

#z^2+z+1=0# means #z=omega# or #omega^2#, where #omega# represents complex cube root of #1+i0#.

From the properties of cube roots of #1#, we know that #1+omega+omega^2=0# as also #1*omega*omega^2=1# or #omega*omega^2=1# i.e. roots of #z^2+z+1=0# are reciprocal of each other.

Hence it does not matter whether we use either of the roots to find value of #(z+1/z)^2+(z^2+1/z^2)^2+(z^3+1/z^3)^2+.....+(z^6+1/z^6)^2# and putting #z=omega# we have

#(omega+1/omega)^2=(omega+omega^2)^2=(-1)^2=1#

and #(z^2+1/z^2)^2=(omega^2+omega)^2=1#

#(z^3+1/z^3)^2=(1+1)^2=4#

#(z^4+1/z^4)^2=(omega+omega^2)^2=(-1)^2=1#

#(z^3+1/z^3)^2=(omega^2+omega)^2=1#

#(z^6+1/z^6)^2=(1+1)^2=4#

Hence adding them all

#(z+1/z)^2+(z^2+1/z^2)^2+(z^3+1/z^3)^2+.....+(z^6+1/z^6)^2=12#

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