Sum of squares of modulus of all the complex numbers z satisfying z=iz^2+z^2-z is equal to

Question

Accepted Answer

Correct answer: 2. (2) z + z = iz^2 + z^2 Consider z = x + iy 2x = (i + 1) (x^2 - y2 + 2xyi) ⇒2x = x2 - y2 - 2xy and x2 - y2 + 2xy = 0 ⇒ 2x = - 4xy ⇒x = 0 or y = -12 Case 1 : x = 0 ⇒ y = 0 here z = 0 Case 2 : y= -12 ⇒4x^2-4x-1=0 (2x-1)^2=2 2x-1= 2 x= 1 22 Here z= 1+ 22- i2 or z= 1- 22- i2 Sum of squares of modulus of z =0+ (1+ 2)^2+14+ (1- 2)^2+14= 84=2

$Sum of squares of modulus of all the complex numbers z satisfying z ˉ = i z^{2} + z^{2} - z is equal to$

Solution

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