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?

(B) $|z| \leq 2$ for all $z \in S$; (C) $\left|z+\frac{1}{2}\right| \geq \frac{1}{2}$ for all $z \in S$

(A) $\left|z+\frac{1}{2}\right| \leq \frac{1}{2}$ for all $z \in S$

(D) The set S has exactly four elements

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Complex Numbers

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?

Answer: (B), (C)

Given

∣ z^{2} + z + 1∣ = 1

, we can write

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

.

Using the triangle inequality:

(z + \frac{1}{2})^{2} + \frac{3}{4} \leq z + \frac{1}{2}^{2} + \frac{3}{4}

1 \leq z + \frac{1}{2}^{2} + \frac{3}{4}

, giving

z + \frac{1}{2}^{2} \geq \frac{1}{4}

, thus

z + \frac{1}{2} \geq \frac{1}{2}

. (C) is TRUE.

Also,

∣ (z^{2} + z) + 1∣ = 1 \geq ∣∣ z^{2} + z ∣ - 1∣

, so

∣ z^{2} + z ∣ \leq 2

.

Since

∣∣ z ∣^{2} - ∣ z ∣∣ \leq ∣ z^{2} + z ∣ \leq 2

, we have

∣ r^{2} - r ∣ \leq 2

where

r = ∣ z ∣

. This gives

r \leq 2

. (B) is TRUE.

(A) is FALSE as we showed

z + \frac{1}{2} \geq \frac{1}{2}

, not

\leq \frac{1}{2}

.

(D) is FALSE because the equation

z^{2} + z + 1 = e^{i θ}

has solutions for all

θ \in R

, so S is infinite.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options