Question

Let C be the set of complex numbers. Prove that the mapping f: C → R given by f (z) = |z|, ∀ z ∈ C, is neither one-one nor onto.

Answer 1

Given, f: C → R such that f (z) = |z|, ∀ z ∈ C

Now, let take z = 6 + 8i

Then,

f (6 + 8i) = |6 + 8i| = √(6² + 8²) = √100 = 10

And, for z = 6 – 8i

f (6 – 8i) = |6 – 8i| = √(6² + 8²) = √100 = 10

Hence, f (z) is many-one.

Also, |z| ≥ 0, ∀ z ∈ C

But the co-domain given is ‘R’

Therefore, f(z) is not onto.