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Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a=z^{2}+z+1$ is real. Then a cannot take the value
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Verified Answer
The correct answer is:
$\frac{3}{4}$
$\because \quad \operatorname{Im}(z) \neq 0$ $\Rightarrow \quad z$ is non-real
and equation $z^{2}+z+(1-a)=0$
will have non-real roots, if $D < 0$
$\Rightarrow \quad 1-4(1-a) < 0$ $\Rightarrow \quad 4 a < 3 \Rightarrow a < \frac{3}{4}$
$\therefore \quad a$ can not take the value $\frac{3}{4}$.
and equation $z^{2}+z+(1-a)=0$
will have non-real roots, if $D < 0$
$\Rightarrow \quad 1-4(1-a) < 0$ $\Rightarrow \quad 4 a < 3 \Rightarrow a < \frac{3}{4}$
$\therefore \quad a$ can not take the value $\frac{3}{4}$.
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