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. $\left|z+\frac{1}{2}\right|<\frac{1}{2},\forall z\in S$
• B. $\left|z\right|<2,\forall z\in S$
• C. $\left|z+\frac{1}{2}\right|\ge \frac{1}{2},\forall z\in S$
• D. The $S$ has exactly four elements

Answer 1

Answer: (B), (C)

$\left|z+\frac{1}{2}\right|\ge \frac{1}{2},\forall z\in S$

Explanation For The Correct Option:

Determining the correct statement

The given expression $|z^2+z+1|=1$

$$\begin{gathered} \Rightarrow |z^2+z+\frac{1}{4}-\frac{1}{4}+1|=1 \\ \Rightarrow \left|{\left(z+\frac{1}{2}\right)}^2+\frac{3}{4}\right|=1 \\ \Rightarrow 1\le {\left|z+\frac{1}{2}\right|}^2+\frac{3}{4}\left[\because \left|a+b\right|\le \left|a\right|+\left|b\right|\right] \\ \Rightarrow {\left|z+\frac{1}{2}\right|}^2\ge \frac{1}{4} \\ \Rightarrow \left|z+\frac{1}{2}\right|\le -\frac{1}{2}Or\left|z+\frac{1}{2}\right|\ge \frac{1}{2}.....\left(i\right) \end{gathered}$$

Now again proceeding

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

$$\begin{gathered} \Rightarrow \left|{\left|z+\frac{1}{2}\right|}^2-\frac{3}{4}\right|\le 1 \\ \Rightarrow -1\le {\left|z+\frac{1}{2}\right|}^2-\frac{3}{4}\le 1 \\ \Rightarrow -\frac{1}{4}\le {\left|z+\frac{1}{2}\right|}^2\le \frac{7}{4} \\ \Rightarrow {\left|z+\frac{1}{2}\right|}^{}\le \frac{\sqrt{7}}{2} \\ \Rightarrow \left|z\right|\le \frac{\sqrt{7}-1}{2}<\frac{4}{2} \\ \Rightarrow \left|z\right|<2.....\left(ii\right) \end{gathered}$$

Hence from equations $\left(i\right)\&\left(ii\right)$ we can say that the correct answer are option(s) (B) and (C).