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Let $S$ be the set of all complex numbers $z$ satisfying $\left|z^2+z+1\right|=1$. Then which of the following statements is/are $TRUE$?
$(A)$ $\left|z+\frac{1}{2}\right| \leq \frac{1}{2}$ for all $z \in S$ $(B)$ $|z| \leq 2$ for all $z \in S$
$(C)$ $\left|z+\frac{1}{2}\right| \geq \frac{1}{2}$ for all $z \in S$ $(D)$ The set $S$ has exactly four elements
$A,C$
$B,C$
$B,D$
$A,D$
Solution
$\left|z^2+z+1\right|=1$
$\Rightarrow \left|\left(z+\frac{1}{2}\right)^2+\frac{3}{4}\right|=1$
$\Rightarrow \left|\left(z+\frac{1}{2}\right)^2+\frac{3}{4}\right| \leq\left|z+\frac{1}{2}\right|^2+\frac{3}{4}$
$\Rightarrow 1 \leq\left|z+\frac{1}{2}\right|^2+\frac{3}{4} \Rightarrow\left|\left(z+\frac{1}{2}\right)\right|^2 \geq \frac{1}{4}$
$\Rightarrow \left|z+\frac{1}{2}\right| \geq \frac{1}{2}$
$\text { also } \left|\left(z^2+z\right)+1\right|=1 \geq|| z^2+z|-1|$
$\Rightarrow \left|z^2+z\right|-1 \leq 1$
$\Rightarrow \left|z^2+z\right| \leq 2$
$\Rightarrow \left\|z^2|-| z\right\| \leq\left|z^2+z\right| \leq 2$
$\Rightarrow \left|r^2-r\right| \leq 2$
$\Rightarrow \quad r =|z| \leq 2 ; \forall z \in S$
Also we can always find root of the equation $z^2+z+1=e^{i \theta} ; \forall \theta \in R$
Hence set ' $S$ ' is infinite
