The common roots of the equations z3+(1+i) z2+(1+i) z+i=0, (where .i=√-1) and z1993+z1994+1=0 are (where ω denotes the complex cube root of unity)

Question

The common roots of the equations z3+(1+i) z2+(1+i) z+i=0, (where .i=√-1) and z1993+z1994+1=0 are (where ω denotes the complex cube root of unity) (A) 1 (B) ω (C) ω2 (D) ω981

Tardigrade Admin · Accepted Answer

z3+(1+i) z2+(1+i) z+i=0 ⇒ (z+i)(z2+z+1)=0 ⇒ (z+i)(z-ω)(z-ω2)=0 ⇒ z=-i, ω, ω2 Now ω, and ω2 satisfies the equation z1993+z1994+1=0 So ω and ω2 are common roots

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

1

$ω$

$ω^{2}$

$ω^{981}$

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

Q. The common roots of the equations z3+(1+i)z2+(1+i)z+i=0, (where i=−1​) and z1993+z1994+1=0 are (where ω denotes the complex cube root of unity)

1

ω

ω2

ω981

z3+(1+i)z2+(1+i)z+i=0 ⇒(z+i)(z2+z+1)=0 ⇒(z+i)(z−ω)(z−ω2)=0 ⇒z=−i,ω,ω2 Now ω, and ω2 satisfies the equation z1993+z1994+1=0 So ω and ω2 are common roots

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

$ω$

$ω^{2}$

$ω^{981}$

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