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JEE Advanced Mathematics Complex Number 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Mathematics Question (2020) — Solution

Question

For a complex number z , let  Re z denote the real part of  z . Let S be the set of all complex numbers z satisfying z 4 - | z | 4 = 4 i z 2 ,  where i = - 1 . Then the minimum possible value of z 1 - z 2 2 , where z 1 , z 2 ∈ S  with  Re z 1 > 0  and  Re z 2 < 0 , is _______

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

z 4 - | z | 4 = 4 i z 2 ⇒ z 4 - z z ¯ 2 = 4 i z 2 ⇒    z 2 z 2 - z ¯ 2 = 4 i z 2    ⇒    z 2 - z ¯ 2 = 4 i ⇒    z + z ¯ z - z ¯ = 4 i ⇒    z + z ¯ 2 z - z ¯ 2 i = 1    ⇒    x y = 1 for  z 1   &   z 2    ⇒    x 1 y 1 = 1 and  x 2 y 2 = 1 x 1   &   x 2 are of opposite sign, similarly  y 1  & y 2 are of opposite sign ⇒    x 1 > 0 , y 1 > 0 ,   x 2 < 0 , y 2 < 0 Now    z 1 - z 2 2 = x 1 - x 2 2 + y 1 - y 2 2 = x 1 2 + x 2 2 + y 1 2 + y 2 2 - 2 x 1 x 2 - 2 y 1 y 2 = x 1 2 + x 2 2 + y 1 2 + y 2 2 + x 1 - x 2 + x 1 - x 2 + y 1 - y 2 + y 1 - y 2 ≥ 8 x 1 2 · x 2 2 · y 1 2 · y 2 2 · x 1 - x 2 · x 1 - x 2 · y 1 - y 2 · y 1 - y 2 1 / 8 ≥ 8 x 1 y 1 4 x 2 y 2 4 1 / 8 ≥ 8 .

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Related: Mathematics — Complex Number · All PYQ Banks