Sum of the common roots of z2006-z100-1-0and z3-2 z2-2 z-1-0 isA- 0B- -1C- 1D- 2

(z^3+1)+(2z^2+2z)=0 ⇒(z+1)(z^2 z+1)+2z(z+1)=0 ⇒ (z+1)(z^2+z+1)=0 ⇒ z= 1, w, w^2 Out of these w and w^2 satisfy z^2006+z^100+1=0 ∴Sum of common roots = w+w^2= 1 ...

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Standard XII

Mathematics

Cube Root of a Complex Number

Sum of the co...

Question

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

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Solution

The correct option is B
-1

$(z^{3} + 1) + (2 z^{2} + 2 z) = 0$
$\Rightarrow (z + 1) (z^{2} - z + 1) + 2 z (z + 1) = 0 \Rightarrow (z + 1) (z^{2} + z + 1) = 0 \Rightarrow z = - 1, w, w^{2}$
Out of these $w$ and $w^{2}$ satisfy
$z^{2006} + z^{100} + 1 = 0$
$∴ Sum of common roots = w + w^{2} = - 1$

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Sum of the common roots of z2006+z100+1=0 and z3+2z2+2z+1=0 is

The correct option is B -1 (z3+1)+(2z2+2z)=0 ⇒(z+1)(z2−z+1)+2z(z+1)=0⇒(z+1)(z2+z+1)=0⇒z=−1,w,w2 Out of these w and w2 satisfy z2006+z100+1=0 ∴Sum of common roots =w+w2=−1

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

The correct option is B
-1

$(z^{3} + 1) + (2 z^{2} + 2 z) = 0$
$\Rightarrow (z + 1) (z^{2} - z + 1) + 2 z (z + 1) = 0 \Rightarrow (z + 1) (z^{2} + z + 1) = 0 \Rightarrow z = - 1, w, w^{2}$
Out of these $w$ and $w^{2}$ satisfy
$z^{2006} + z^{100} + 1 = 0$
$∴ Sum of common roots = w + w^{2} = - 1$