Question

Solve the equation |z| = z + 1 + 2i.

Answer 1

Given that:  |z| = z + 1 + 2i

Let z = x + iy

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

Squaring both sides

|z|² = |z + 1|² + 4i² + 4(z + 1)i

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

⇒ 0 = –3 + 2z + 4(z + 1)i

⇒ 3 – 2z – 4(z + 1)i = 0

⇒ 3 – 2(x + yi) – 4[x + yi + 1]i = 0

⇒ 3 – 2x – 2yi – 4xi – 4yi2 – 4i = 0

⇒ 3 – 2x + 4y – 2yi – 4i – 4xi = 0

⇒ (3 – 2x + 4y) – i(2y + 4x + 4) = 0

⇒ 3 – 2x + 4y = 0 

⇒ 2x – 4y = 3  .....(i)

And 4x + 2y + 4 = 0 

⇒ 2x + y = –2  .....(ii)

Solving equation (i) and (ii), we get

y = –1 and x = `-1/2`

Hence, the value of z = x + yi = `(- 1/2 - i)`.