Verified Solution • 2025–26 Session moderate

QJEEAdv20Mab2f3: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Mab2f3 Solution: Let S be the set of all complex numbers z satisfying

Let S be the set of all complex numbers z
satisfying |z2 + z + 1| = 1. Then which of the following statements is/are TRUE?

Multiple Choice Options

  • A ∣z+12∣\left| {z + {1 \over 2}} \right| ≤\le 12{{1 \over 2}} for all z ∈\in S
  • B |z| ≤\le 2 for all z ∈\in S
  • C ∣z+12∣ ≥12\left| {z + {1 \over 2}} \right|\, \ge {1 \over 2} for all z ∈\in S
  • D The set S has exactly four elements

Step-by-Step Verified Solution

It is given that the complex number satisfying

∣z2+z+1∣=1⇒∣(z+12)2+34∣=1|{z^2} + z + 1| = 1 \Rightarrow \left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right| = 1

∵\because ∣z1−z2∣ ≥ ∣∣z1∣−∣z2∣∣|{z_1} - {z_2}|\, \ge \,||{z_1}| - |{z_2}||

∴\therefore ∣(z+12)2−(−34)∣ ≥ ∣∣z+12∣2−∣−34∣∣\left| {{{\left( {z + {1 \over 2}} \right)}^2} - \left( { - {3 \over 4}} \right)} \right|\, \ge \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} - \left| { - {3 \over 4}} \right|} \right|

⇒∣∣z+12∣2−34∣ ≤ 1\Rightarrow \left| {{{\left| {z + {1 \over 2}} \right|}^2} - {3 \over 4}} \right|\, \le \,1

⇒−1≤ ∣z+12∣2−34 ≤ 1\Rightarrow - 1 \le \,{\left| {z + {1 \over 2}} \right|^2} - {3 \over 4}\, \le \,1

⇒−14 ≤ ∣z+12∣2 ≤ 74\Rightarrow - {1 \over 4}\, \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}

⇒0≤ ∣z+12∣2 ≤ 74\Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4} { ∵\because |z| ≥\ge 0}

⇒0≤ ∣z+12∣2 ≤ 72\Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{{\sqrt 7 } \over 2} ....(i)

∵\because ⇒ ∣z+12∣ ≤ 72\Rightarrow \,\left| {z + {1 \over 2}} \right|\, \le \,{{\sqrt 7 } \over 2}

∵\because ∣z1+z2∣ ≥ ∣∣z1∣−∣z2∣∣|{z_1} + {z_2}|\, \ge \,||{z_1}| - |{z_2}||

∴\therefore ∣z+12∣ ≥ ∣∣z∣−12∣\left| {z + {1 \over 2}} \right|\, \ge \,\left| {|z| - {1 \over 2}} \right|

⇒∣∣z∣−12∣ ≤ ∣z+12∣≤ 72\Rightarrow \left| {|z| - {1 \over 2}} \right|\, \le \,\left| {z + {1 \over 2}} \right| \le \,{{\sqrt 7 } \over 2}

⇒−72 ≤ ∣z∣−12≤72\Rightarrow - {{\sqrt 7 } \over 2}\, \le \,|z| - {1 \over 2} \le {{\sqrt 7 } \over 2}

⇒1−72≤ ∣z∣ ≤7+12\Rightarrow {{1 - \sqrt 7 } \over 2} \le \,|z|\, \le {{\sqrt 7 + 1} \over 2}

⇒∣z∣ ≤ 1+72\Rightarrow |z|\, \le \,{{1 + \sqrt 7 } \over 2} , ∴\therefore |z| ≤\le 2

∵\because ∣(z+12)2+34∣ ≤ ∣∣z+12∣2+34∣\left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right|\, \le \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right|

⇒1≤∣∣z+12∣2+34∣⇒∣z+12∣2+34≥1\Rightarrow 1 \le \left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right| \Rightarrow {\left| {z + {1 \over 2}} \right|^2} + {3 \over 4} \ge 1

⇒∣z+12∣2≥14⇒∣z+12∣ ≥12\Rightarrow {\left| {z + {1 \over 2}} \right|^2} \ge {1 \over 4} \Rightarrow \left| {z + {1 \over 2}} \right|\, \ge {1 \over 2} ....(ii)

from Eqs. (i) and (ii), we get

12≤∣z+12∣ ≤72{1 \over 2} \le \left| {z + {1 \over 2}} \right|\, \le {{\sqrt 7 } \over 2}

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