Mathematics - Complex Number Question with Solution | TestHub

$z^2+z+1=e^{i\theta }\theta \in (-\pi , \pi ]$

$\Rightarrow z^2+z+1-e^{i\theta }=0$

By quadratic formula, we get

$z=\frac{-1\pm \sqrt{4e^{i\theta }-3}}{2} .............\left(1\right)$

$\Rightarrow z+\frac{1}{2}=\pm \sqrt{\left(4\cos \theta -3\right)+i4\sin \theta }$
$\Rightarrow \left|z+\frac{1}{2}\right|={\left[{\left(4\cos \theta -3\right)}^2+{\left(4\sin \theta \right)}^2\right]}^{\frac{1}{4}}$

$\Rightarrow \left|z+\frac{1}{2}\right|\in \left[1, \sqrt{7}\right]$

Option (C) is correct
By equation $\left(1\right)$,

$\left|2z\right|\le 1+\sqrt{\left|4e^{i\theta }-3\right|}$

$\Rightarrow \left|2Z\right|\le 1+{(25-24\cos \theta )}^{\frac{1}{4}}$

$\Rightarrow \left|2Z\right|\le 1+\sqrt{7}<4$

$\Rightarrow \left|Z\right|\le 2$

Option (B) is correct.