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

Answer 1

Answer: (B), (C)

$\left|z^2+z+1\right|=1$
$\Rightarrow \left|{\left(z+\frac{1}{2}\right)}^2+\frac{3}{4}\right|=1$
$\Rightarrow {\left(z+\frac{1}{2}\right)}^2\left|-\frac{3}{4}\le 1\le \right|z+{\frac{1}{2}|}^2+\frac{3}{4}$
$\Rightarrow \frac{1}{4}\le {\left|z+\frac{1}{2}\right|}^2\le \frac{7}{4}$
$\Rightarrow \frac{1}{2}\le \left|z+\frac{1}{2}\right|\le \frac{\sqrt{7}}{2}$
option (c) is correct
$\because \left|z+\frac{1}{2}\right|\le \frac{\sqrt{7}}{2}$
$\Rightarrow |z|-\frac{1}{2}\le \frac{\sqrt{7}}{2}$
$\Rightarrow |z|\le \frac{\sqrt{7}+1}{2}$
$\Rightarrow |z|\le 2$ 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