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If $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{Z}: \mathrm{x}^{2}-1=0\right\}$ and $\mathrm{B}=\left\{\mathrm{x} \in \mathrm{Z}: \mathrm{x}^{2}+\mathrm{x}+1=0\right\}$, where
Z is set of complex numbers, then what is $A \cap B$ equal to ?
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Z is set of complex numbers, then what is $A \cap B$ equal to ?
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Verified Answer
The correct answer is:
$\left\{\frac{-1+\sqrt{3} \mathrm{i}}{2}, \frac{-1-\sqrt{3} \mathrm{i}}{2}\right\}$
Given, $\mathrm{A}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^{3}-1=0\right\}$
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^{2}+\mathrm{x}+1=0\right\}$
The roots of $\mathrm{x}^{3}-1=0$ are $1, \omega, \omega^{2}$ The roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$ $\therefore A \cap B=\left\{1, \omega, \omega^{2}\right\} \cap\left\{\omega, \omega^{2}\right\}=\left\{\omega, \omega^{2}\right\}$
$=\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}$
$\mathrm{B}=\left\{\mathrm{x} \in \mathrm{z}: \mathrm{x}^{2}+\mathrm{x}+1=0\right\}$
The roots of $\mathrm{x}^{3}-1=0$ are $1, \omega, \omega^{2}$ The roots of $x^{2}+x+1=0$ are $\omega, \omega^{2}$ $\therefore A \cap B=\left\{1, \omega, \omega^{2}\right\} \cap\left\{\omega, \omega^{2}\right\}=\left\{\omega, \omega^{2}\right\}$
$=\left\{\frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}\right\}$
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