Question

If $z$ is 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

$\frac{3}{4}$

Explanation for the correct option.

It is given that the imaginary part of the complex number $z$ is non-zero.

Also, the equation $a=z^2+z+1$ can be written as: $z^2+z+1-a=0$.

Now the quadratic equation $z^2+z+1-a=0$ in $z$ will have non-real solutions because the complex number $z$ has an imaginary part.

A quadratic equation has non-real solutions when its discriminant is less than $0$.

Now, comparing the equation $z^2+z+1-a=0$ with standard quadratic equation $a{z}^2+bz+c=0$ :

$$\begin{gathered} a=1 \\ b=1 \\ c=1-a \end{gathered}$$

So the discriminant $[{\mathrm{b}}^2-4\mathrm{ac}]$ is less than $0$ and so the inequality is:

$$\begin{gathered} b^2-4ac<0 \\ \Rightarrow 1-4\times 1\times \left(1-a\right)<0 \\ \Rightarrow 1-4+4a<0 \\ \Rightarrow -3+4a<0 \\ \Rightarrow 4a<3 \\ \Rightarrow a<\frac{3}{4} \end{gathered}$$

So, the real number $a$ cannot take the value of $\frac{3}{4}$.

Hence, the correct option is D.