Question

The set of points on the complex plane such that $z^2+z+1$ is real and positive (where $z=x+iy,x,yϵR)$ is

• A. Complete real axis only
• B. Complete real axis or all points on the line $2x+1=0$
• C. Complete real axis or a line segment joining points $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ & $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$ excluding both
• D. Complete real axis or set of points lying inside the rectangle formed by the lines.
  $2x+1=0;2x-1=0;2y-\sqrt{3}=0$ & $2y+\sqrt{3}=0$

Answer 1

The correct option is C Complete real axis or a line segment joining points $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ & $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$ excluding both

$z^2+z+1$ is real so
$z^2+z+1={\overline{z}}^2+\overline{z}+1$
$z^2{\overline{z}}^2+z-\overline{z}=0$
$\left(z\overline{z}\right)(z+\overline{z}+1)=0$
either $z=z$ or $z+\overline{z}+1=0$
$\Rightarrow lm\left(z\right)=0$ Let $z=\alpha +i\beta$
$\Rightarrow$ $\alpha +i\beta +\alpha -i\beta +1=0$
$\Rightarrow 2\alpha +1=0$
$\Rightarrow \alpha =-\frac{1}{2}$
Also $(\alpha +i\beta {)}^2+(\alpha +i\beta )+1>0$
${\alpha }^2+\alpha +1-{\beta }^2+i(2\alpha \beta +\beta )>0$
if $\alpha =-1/2$ then
$\frac{1}{4}-\frac{1}{2}+1-{\beta }^2>0$
$\Rightarrow {\beta }^2-\frac{3}{4}<0\Rightarrow -\frac{\sqrt{3}}{2}<\beta <\frac{\sqrt{3}}{2}$