Question

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

Answer 1

Answer: (B), (C)

$\left|z^{2}+z+1\right|=1$
$\Rightarrow \left|\left(z+\frac{1}{2}\right)^{2}+\frac{3}{4}\right|=1$
$\Rightarrow \left(z+\frac{1}{2}\right)^{2}\left|-\frac{3}{4} \leq 1 \leq\right| z+\left.\frac{1}{2}\right|^{2}+\frac{3}{4}$
$\Rightarrow \frac{1}{4} \leq\left|z+\frac{1}{2}\right|^{2} \leq \frac{7}{4}$
$\Rightarrow \frac{1}{2} \leq\left|z+\frac{1}{2}\right| \leq \frac{\sqrt{7}}{2}$
option (c) is correct
$\because\left|z+\frac{1}{2}\right| \leq \frac{\sqrt{7}}{2}$
$\Rightarrow |z|-\frac{1}{2} \leq \frac{\sqrt{7}}{2}$
$ \Rightarrow |z| \leq \frac{\sqrt{7}+1}{2}$
$\Rightarrow |z| \leq 2$ option (B) is correct
Clearly $z=0,-1, i,-i,-1 \pm i$ a
re satisfying the given equation.
So we have more than $4$ solutions