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JEE Main202325 Jan 2023Evening ShiftMathematicsComplex NumberActual

Let z be a complex number such that z - 2 i z + i = 2 , z ≠ - i . Then z lies on the circle of radius 2 and centre

Options

  1. A( 2 , 0 )
  2. B( 0 , 2 )
  3. C( 0 , 0 )
  4. D( 0 , - 2 )

Correct answer

D. ( 0 , - 2 )

Step-by-step solution

Given equation is z - 2 i z + i = 2 ,   z ≠ - i On simplifying the equation, we get ( z - 2 i ) ( z ¯ + 2 i ) = 4 ( z + i ) ( z ¯ - i ) ⇒ z z ¯ + 4 + 2 i ( z - z ¯ ) = 4 ( z z ¯ + 1 + i ( z ¯ - z ) ) ⇒ 3 z z ¯ - 6 i ( z - z ¯ ) = 0 Now putting the value of z = x + i y   &   z ¯ = x - i y we get, ⇒ x 2 + y 2 - 2 i ( 2 i y ) = 0 ⇒ x 2 + y 2 + 4 y = 0 On comparing the above equation with the general equation of the circle i.e

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