Question

If f : C → C, f(x + iy) = (x – iy), then prove that f is an one-one onto function.

Answer 1

Given : f : C → C and f(x + iy) = x – iy

where C is set of complex numbers.

Let x₁ + iy₁ and x₂ + iy₂ ∈ C thus.

f(x₁ + iy₁) = f (x₂ + iy₂)

⇒ x₁ – iy₁ = x₂ – iy₂

So, f is one-one function.

Range of f = {x – iy : x + iy ∈ C} = C (co-domain)

f is onto function.

Hence, f is one-one, onto function.

Hence Proved.