Question

Sum of the common roots of $z^{2006}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is B $-1$

$z^3+2z^2+2z+1=0$ ....(1)
$\Rightarrow (z+1)(z+w)(z+w^2)=0$
Therefore,$-1,-w,-w^2$ are the roots if the equation (1)
where, $w$ is the cube root of unity.
$\Rightarrow w^3=1\&1+w+w^2=0$
Consider, $f\left(z\right)=z^{2006}+z^{100}+1$
Now, $f\left(w\right)=w^{2006}+w^{100}+1={\left(w^3\right)}^{668}{w}^2+{\left(w^3\right)}^{33}w+1$
$\Rightarrow f\left(w\right)=1+w+w^2=0$
and $f\left(w^2\right)={\left(w^2\right)}^{2006}+{\left(w^2\right)}^{100}+1={\left(w^3\right)}^{1337}w+{\left(w^3\right)}^{66}{w}^2+1$
$\Rightarrow f\left(w^2\right)=1+w+w^2=0$
Since, $f\left(w\right)=0$ and $f\left(w^2\right)=0$
Therefore, $w$ and $w^2$ are the roots of the equation $z^{2006}+z^{100}+1$.
and hence, $w$ and $w^2$ are common root
$\Rightarrow w+w^2=-1$
Ans: B