Q28 / 60 · JEE Advanced 2012 (Paper 1) · Mathematics

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

A
$-1$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
Correct answer
D
$\frac{3}{4}$

Solution

$\because \quad \operatorname{Im}(z) \neq 0$ $\Rightarrow \quad z$ 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 $\frac{3}{4}$.

Hence, the correct answer is option: 'D'.

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