Question

If $z^2+z+1=0,$ where z is a complex number, then the value of $(z+\frac{1}{z}{)}^2+(z^2+\frac{1}{z^2}{)}^2+(z^3+\frac{1}{z^3}{)}^2+\dots \dots .+(z^6+\frac{1}{z^6}{)}^2$

• A. 18
• B. 54
• C. 6
• D. 12

Answer 1

The correct option is D 12

$z^2+z+1=0$

divide by $z,$

$\Rightarrow z+\frac{1}{z}=-1$

squaring $z^2+\frac{1}{z^2}+2=1$

$\Rightarrow z^2+\frac{1}{z^2}=-1$

$z^3+\frac{1}{z^3}=(z+\frac{1}{z})(z^2+\frac{1}{z^2}-1)=(-1)(-1-1)=2$

squaring $(z^2+\frac{1}{z^2}=-1)\Rightarrow z^4+\frac{1}{z^4}=-1$

$z^5+\frac{1}{z^5}+z^3+\frac{1}{z^3}=(z^4+\frac{1}{z^4})(z+\frac{1}{z})$

$\therefore z^5+\frac{1}{z^5}=(-1)(-1)-2=-1$

squaring $z^3+\frac{1}{z^3}=2\Rightarrow z^6+\frac{1}{z^6}=2$

now $(z+\frac{1}{z}{)}^2+(z^2+\frac{1}{z^2}{)}^2+(z^3+\frac{1}{z^3}{)}^2+---+(z^6+\frac{1}{z^6}{)}^2$

$\Rightarrow 1+1+4+1+1+4=12$