Question

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. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

If $a{x}^2+bx+c=0$ has roots $\alpha ,\beta$ then
$\alpha ,\beta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
For roots to be $b^2-4ac\ge 0$.
Description of Situation As imaginary part of z = x+iy is non- zero.
$\Rightarrow y\ne 0$
Method 1: Let $z=x+iy$
$\therefore a=(x+iy{)}^2+(x+iy)+1$
$\Rightarrow (x^2-y^2+x+1-a)+i(2xy+y)=0\Rightarrow (x^2-y^2+x+1-a)+iy(2x+1)=0$, ...(1)
It is purely real, if y(2x+1) = 0
But imaginary part of z. i.e y is non-zero.
$\Rightarrow 2x+1=0orx=-\frac{1}{2}$
From Eq. (1), $\frac{1}{4}-y^2-\frac{1}{2}+1-a=0$
$\Rightarrow a=-y^2+\frac{3}{4}\Rightarrow a<\frac{3}{4}$
Method 2: Here, $z^2+z+(1-a)=0$
$\therefore z=\frac{-1\pm \sqrt{1-4(1-\alpha )}}{2}\Rightarrow z=\frac{-1\pm \sqrt{\left(4a\right)-3}}{2}$
For z do not have real roots, $4a-3<0\Rightarrow a<\frac{3}{4}$