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

$S = \{\, z \in \mathbb{C} : |z^2 + z + 1| = 1 \,\}$

$\text{Check statements (1), (2), (3), (4) for every } z \in S$

$w = z + \tfrac{1}{2} \quad\Rightarrow\quad z = w - \tfrac{1}{2}$

$\left(w - \tfrac{1}{2}\right)^2 + \left(w - \tfrac{1}{2}\right) + 1 = w^2 + \tfrac{3}{4}$

$\left|w^2 + \tfrac{3}{4}\right| = 1$

$w^2 + \tfrac{3}{4} \ \text{on the unit circle}$

$\Rightarrow w^2 \ \text{on circle centred at } -\tfrac{3}{4},\ \text{radius } 1$

$\tfrac{1}{4} \le |w^2| \le \tfrac{7}{4} \quad\Rightarrow\quad \tfrac{1}{2} \le |w| \le \tfrac{\sqrt{7}}{2}$

$|z + \tfrac{1}{2}| = |w| \ge \tfrac{1}{2}$

$\therefore \text{(3) is TRUE}$

$|w| = \tfrac{1}{2} \text{ only when } w = \pm\tfrac{1}{2} \Rightarrow z = 0,\, -1$

$\therefore \text{(1) is FALSE}$

$\text{If } |z| > 2:\ |z^2+z+1| \ge |z|^2 - |z| - 1 > 1$

$\Rightarrow \text{contradiction, so } |z| \le 2$

$\therefore \text{(2) is TRUE}$

$w^2 \text{ sweeps a whole circle} \Rightarrow S \text{ is infinite (a curve)}$

$\therefore \text{(4) is FALSE}$

$\boxed{\text{Correct statements: (2) and (3)}}$