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
⇒ x = \(-\frac{1}{2}\)
Hence, the value of z = x + iy
= \(-\frac{1}{2}\) + i(-1)
z = \(-\frac{1}{2}\) - i