Question

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?

• A. $|z+\frac{1}{2}|\le \frac{1}{2}$ for all $z\in S$
• B. $|z|\le 2$ for all $z\in S$
• C. $|z+\frac{1}{2}|\ge \frac{1}{2}$ for all $z\in S$
• D. The set $S$ has exactly four elements

Answer 1

Answer: (B), (C)

$|z+\frac{1}{2}|\ge \frac{1}{2}$ for all $z\in S$

$|z^2+z+1|=1$
$\Rightarrow |{(z+\frac{1}{2})}^2+\frac{3}{4}|=1$
$\Rightarrow |{(z+\frac{1}{2})}^2+\frac{3}{4}|\le {|z+\frac{1}{2}|}^2+\frac{3}{4}$
$\Rightarrow 1\le {|z+\frac{1}{2}|}^2+\frac{3}{4}\Rightarrow {\left|(z+\frac{1}{2})\right|}^2\ge \frac{1}{4}$
$\Rightarrow |z+\frac{1}{2}|\ge \frac{1}{2}$ option $c$ is correct
Also
$|(z^2+z)+1|=1\ge ||z^2+z|-1|$
$\Rightarrow |z^2+z|-1\le 1$
$\Rightarrow |z^2+z|\le 2$
$\Rightarrow ||z^2|-|z||\le |z^2+z|\le 2$
$\Rightarrow |r^2-r|\le 2$
$\Rightarrow r=|z|\le 2;\forall z\in S\because r\ge 0$

Also we can always find root of the equation $z^2+z+1=e^{i\theta },\forall \theta \in R$
Hence set ${}^{\prime }{S}^{\prime }$ is infinite.