Question

The common roots of the equations $z^3+(1+i) z^2+(1+i) z+i=0$, (where $\left.i=\sqrt{-1}\right)$ and $z^{1993}+z^{1994}+1=0$ are (where $\omega$ denotes the complex cube root of unity)

• A. 1
• B. $\omega$
• C. $\omega^2$
• D. $\omega^{981}$

Answer 1

Answer: (B), (C)

$z^3+(1+i) z^2+(1+i) z+i=0$
$\Rightarrow (z+i)\left(z^2+z+1\right)=0 $
$\Rightarrow (z+i)(z-\omega)\left(z-\omega^2\right)=0$
$\Rightarrow z=-i, \omega, \omega^2$
Now $\omega$, and $\omega^2$ satisfies the equation
$z^{1993}+z^{1994}+1=0$
So $\omega$ and $\omega^2$ are common roots