$\text{ Let }\mathrm{S}\text{ be the set of all complex numbers }\mathrm{z}\text{ satisfying}\left|{\mathrm{z}}^2+\mathrm{z}+1\right|=1\text{ . Then which of the following statements is/are TRUE ? }$

$\begin{array}{l}\left|z^2+z+1\right|=1\\ \Rightarrow \left|{\left(z+\frac{1}{2}\right)}^2+\frac{3}{4}\right|=1\\ \Rightarrow 1=\left|{\left(z+\frac{1}{2}\right)}^2+\frac{3}{4}\right|\le {\left|z+\frac{1}{2}\right|}^2+\frac{3}{4}\\ \Rightarrow 1\le {\left|z+\frac{1}{2}\right|}^2+\frac{3}{4}\Rightarrow {\left|\left(z+\frac{1}{2}\right)\right|}^2\ge \frac{1}{4}\\ \Rightarrow \left|z+\frac{1}{2}\right|\ge \frac{1}{2}\\ \text{ also }\left|\left(z^2+z\right)+1\right|=1\ge \parallel z^2+z|-1| \left(\sin ce\left|\left|z_1\right|-\left|z_2\right|\right|\le \left|z_1+z_2\right| \right)\\ \Rightarrow \left|z^2+z\right|-1\le 1\\ \Rightarrow \left|z^2+z\right|\le 2\\ \Rightarrow \parallel z^2|-|z\left|\right|\le \left|z^2+z\right|\le 2\\ \Rightarrow \left|r^2-r\right|\le 2 \left(where r=\left|z\right|\right)\end{array}$

 $\Rightarrow r^2-r\le 2\Rightarrow r^2-r-2\le 0\Rightarrow \left(r-2\right)\left(r+1\right)\le 0\Rightarrow \left(r-2\right)\le 0$

$\begin{array}{l}\Rightarrow \mathrm{r}=\left|\mathrm{z}\right|\le 2;\mathrm{\forall }\mathrm{z}\in \mathrm{S}\end{array}$

$\begin{array}{l}\text{ Also we can always find root of the equation }{\mathrm{z}}^2+\mathrm{z}+1={\mathrm{e}}^{\mathrm{i}\theta };\mathrm{\forall }\theta \in \mathrm{R}\\ \text{ Hence set 'S' is infinite }\end{array}$