If |z + 1| = z + 2 (1 + i), then find z.

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

Putting z = x + iy, we get

⇒ |x + iy + 1| = x + iy + 2 (1 + i)

We know that

Comparing real and imaginary parts,

And 0 = y + 2

⇒ y = -2

Putting this value of y in ,

⇒ (x + 1)² + (-2)² = (x + 2)²

⇒ x² + 2x + 1 + 4 = x² + 4x + 4

⇒ 2x = 1

∴ x = 1/2

∴ z = x + iy

= 1/2 – 2i