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

According to the question,

We have,

|z| = z + 1 + 2i

Substituting z = x + iy, we get,

⇒ |x + iy| = x + iy + 1 + 2i

We know that,

|z| = √(x2 + y2)

√(x2 + y2) = (x + 1) + i(y + 2)

Comparing real and imaginary parts,

We get,

√(x2 + y2) = (x + 1)

And 0 = y + 2

⇒ y = -2

Substituting the value of y in √(x2 + y2) = (x + 1),

We get,

⇒ x2 + (-2)2 = (x + 1)2

⇒ x2 + 4 = x2 + 2x + 1

Hence, x = 3/2

Hence, z = x + iy

= 3/2 – 2i