Let $z$ be a complex number such that the imaginary part of $z$ is nonzero and $a=z^{2}+z+1$ is real

Question

Let $z$ be a complex number such that the imaginary part of $z$ is nonzero and $a=z^{2}+z+1$ is real. Then $a$ cannot take the value\

Professor PC Thomas Classes · Answer

(d) : Let $z=x+i y$ Then $z^{2}+z+1=a$\ $\Rightarrow \quad(x+i y)^{2}+(x+i y)+1=a$\ $\Rightarrow \quad x^{2}-y^{2}+i(2 x y+y)+x+1=a$\ $\Rightarrow \quad\left(x^{2}-y^{2}+x+1ight)+i y(2 x+1)-a=0$\ Now $y=0$ or $x=-\frac{1}{2}$\ For imaginary part to be zero .\ $\Rightarrow \quad x^{2}+x+1-a=0 \Rightarrow D \geq 0$\ $\Rightarrow \quad 1-4(1-a) \geq 0$\ $\Rightarrow-3+4 a \geq 0 \Rightarrow 4 a \geq 3 \Rightarrow a \geq \frac{3}{4}$\ Similarly one can investigate the other possibility.\