Given: |z| = z + 1 + 2i

Consider,

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

Squaring both the sides, we get

|z|² = [(z + 1) + (2i)]²

⇒ |z|² = |z + 1|² + 4i² + 2(2i)(z + 1)

⇒ |z|² = |z|² + 1 + 2z + 4(-1) + 4i(z + 1)

⇒ 0 = 1 + 2z – 4 + 4i(z + 1)

⇒ 2z – 3 + 4i(z + 1) = 0

Let z = x + iy

⇒ 2(x + iy) – 3 + 4i(x + iy + 1) = 0

⇒ 2x + 2iy – 3 + 4ix + 4i²y + 4i = 0

⇒ 2x + 2iy – 3 + 4ix + 4(-1)y + 4i = 0

⇒ 2x – 3 – 4y + i(4x + 2y + 4) = 0

Comparing the real part, we get

2x – 3 – 4y = 0

⇒ 2x – 4y = 3 …(i)

Comparing the imaginary part, we get

4x + 2y + 4 = 0

⇒ 2x + y + 2 = 0

⇒ 2x + y = -2 …(ii)

Subtracting eq. (ii) from (i), we get

2x – 4y – (2x + y) = 3 – (-2)

⇒ 2x – 4y – 2x – y = 3 + 2

⇒ -5y = 5

⇒ y = -1

Putting the value of y = -1 in eq. (i), we get

2x – 4(-1) = 3

⇒ 2x + 4 = 3

⇒ 2x = 3 – 4

⇒ 2x = - 1

Hence, the value of z = x + iy