If |z + 1| = z + 2 (1 + i),

According to the question,

We have,

|z + 1| = z + 2 (1 + i)

Substituting z = x + iy, we get,

⇒ |x + iy + 1| = x + iy + 2 (1 + i)

We know,

|z| = √(x2 + y2)

√((x + 1)2 + y2) = (x + 2) + i(y + 1)

Comparing real and imaginary parts,

⇒ √((x + 1)2 + y2) = x + 2

And 0 = y + 2

⇒ y = -2

Substituting the value of y in √((x + 1)2 + y2) = x + 2,

⇒ (x + 1)2 + (-2)2 = (x + 2)2

⇒ x2 + 2x + 1 + 4 = x2 + 4x + 4

⇒ 2x = 1

Hence, x = ½

Hence, z = x + iy

= ½ – 2i