Question

If $z^2+z+1=0,\text{then}\sum _{r=1}^6{(z_1+\frac{1}{z_2})}^2$ is equal to

• A. 6
• B. 12
• C. 18
• D. 24

Answer 1

Answer: (A)

6

Given: $z^2+z+1=0$

$z=\frac{-1\pm \sqrt{1-4}}{2}$

$=\frac{-1\pm \sqrt{3}i}{2}$

$z_1=\frac{-1+\sqrt{3}i}{2}$

$z_2=\frac{-1-\sqrt{3}i}{2}$

$\frac{1}{z_2}=\frac{-2}{+\sqrt{3}i}\frac{(1-\sqrt{3}i)}{(1-\sqrt{3}i)}=\frac{-2(1-\sqrt{3}i)}{4}$

$\frac{1}{z_2}=\frac{-1+\sqrt{3}i}{2}=z_1$

$\Rightarrow z_1+\frac{1}{z_1}=z_1+z_2=-1$

$\Rightarrow {(z_1+\frac{1}{z_1})}^2=1$

$\Rightarrow \sum _{r=1}^6{(z_1+\frac{1}{z_1})}^2=\sum _{r=1}^{6}1=6$