Question

The common roots of the equations
$z^3+(1+i)z^2+(1+i)z+i=0,$ (where $i=\sqrt{-1}$) and $z^{1993}+z^{1994}+1=0$ are

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

Answer 1

Answer: (B), (C)

The correct options are
B

$\omega$

C

${\omega }^2$

$z^3+(1+i)z^2+(1+i)z+i=0$

$\Rightarrow z^2(z+i)+z(z+i)+(z+i)=0$

$\Rightarrow (z+i)(z^2+z+1)=0$

$\Rightarrow (z+i)(z-\omega )(z-{\omega }^2)=0$

$\therefore z=-i,\omega ,{\omega }^2$

Now in $z^{1993}+z^{1994}+1$ put $z=-i,\omega ,{\omega }^2$

then $(-i{)}^{1993}+(-i{)}^{1994}+1=-i-1+1=-i\ne 0$

and $(\omega {)}^{1993}+{\omega }^{1994}+1=\omega +{\omega }^2+1=0$

and $({\omega }^2{)}^{1993}+({\omega }^2{)}^{1994}+1={\omega }^2+\omega +1=0$