Let S be the set of all complex numbers z satisfying ≤ft|z²+z+1|=1. Then which of the following statements is/are TRUE? [JEE Adv 2020 P1]

≤ft|z+1/2| ≤ 1/2 for all z S

≤ft|z+1/2| ≥ 1/2 for all z S

The set S has exactly four elements

Test Index

Complex Numbers

Section: Mathematics

Question : 13 of 36

Marks: +1, -0

Let

S

z

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

Solution:

\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} \right| \le 1

\le \left|z+\frac{1}{2}\right|^{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 \left|z\right|-\frac{1}{2} \le \frac{\sqrt{7}}{2} \Rightarrow \left|z\right| \le \frac{\sqrt{7}+1}{2}

\Rightarrow \left|z\right| \le 2

option (B) is correct

Clearly

z=0,-1,i,-i,-1\pm i

are satisfying the given equation. So we have more than 4 solutions

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