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?

The set S has exactly four elements

Mathematics · Complex Numbers

JEE Advanced 2020 — Paper 1 — Question 37

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?

Option A:
$\left|z+\frac{1}{2}\right| \leq \frac{1}{2}$ for all $z \in S$
Option B:
$|\mathrm{z}| \leq 2$ for all $\mathrm{z} \in \mathrm{S}$
Correct
Option C:
$\left|z+\frac{1}{2}\right| \geq \frac{1}{2}$ for all $z \in S$
Correct
Option D:
The set S has exactly four elements

$\left|z^{2}+z+1\right|=1$

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

Using $\Delta$ inequality

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

$\left|z+\frac{1}{2}\right|^{2} \geq \frac{1}{4}$

\begin{gathered} \left|z+\frac{1}{2}\right| \geq \frac{1}{2} …(1)\end{gathered}

$\left|\left|z^{2}+z\right|-1\right| \leq 1$

$0 \leq\left|z^{2}+z\right| \leq 2 \Rightarrow\left|z^{2}+z\right| \leq 2$

$|z|^{2}-|z| \leq 2 \Rightarrow|z|^{2}-|z|-2 \leq 0$

$(|z|-2)(|z|+1) \leq 0$

$|z| \leq 2$

Answer key and solution verified before publishing.

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