Question

The common roots of the equation $z^3+(1+i)z^2+(1+i)z+i=0$, (where $i=\sqrt{-1})$ 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)

$\omega$
${\omega }^2$

Simplifying the first equation, we get
$(z+i)(z^2+z+1)=0$
$z=-i,w,w^2$
Clearly out of the three only
$w,w^2$ satisfies
$z^{1993}+z^{1994}+1=0$
$w^{1993}+w^{1994}+1$
$=w^{1993}(1+w)+1$
$=w^{1993}(-w^2)+1$
$=-w^{1995}+1$
$=-(w^3{)}^{665}+1$
$=-1+1$
$=0$
Similarly for $w^2$.