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

Answer 1

Answer: (B), (C)

$|z|\le 2$ for all $zϵS$
$|z+\frac{1}{2}|\ge \frac{1}{2}$ for all $zϵS$

$|z^2+z+1|=1=||{(z+\frac{1}{2})}^2+\frac{3}{4}|$

$1\le {|z+\frac{1}{2}|}^2+\frac{3}{4}$

${|z+\frac{1}{2}|}^2\ge \frac{1}{4}$

$|z+\frac{1}{2}|\le -\frac{1}{2}$ or $|z+\frac{1}{2}|\ge \frac{1}{2}$ ....(C)

Now

$|{(z+\frac{1}{2})}^2+\frac{3}{4}|=|z^2+z+1|=1$

$|{|z+\frac{1}{2}|}^2-\frac{3}{4}|\le 1$
$-1\le {|z+\frac{1}{2}|}^2-\frac{3}{4}\le 1$

$\frac{-1}{4}\le {|z+\frac{1}{2}|}^2\le \frac{7}{4}$

$|z+\frac{1}{2}|\le \frac{\sqrt{7}}{2}$

$|z|\le \frac{\sqrt{7}}{2}-1$

$|z|\le \frac{\sqrt{7}-1}{2}\le \frac{4}{2}$

$|z|\le 2$ ...(B)