Answer :

We have,

f : C → R given by f (z) = |z|, ∀ z ∈ C

In order to prove that f is one-one, it is sufficient to prove that f(z_{1})=f(z_{2}) ⇒ z_{1}=z_{2}∀ z_{1}, z_{2} ∈ C .

Let z_{1} = 2+3i and z_{2} = 2-3i are two distinct complex numbers.

Now,

f(z_{1}) = |z_{1}| = |2+3i | = = √ 13

f(z_{2}) = |z_{2}| = |2-3i | = = √ 13

here, we observe that f(z_{1}) = f(z_{2}) but z_{1} ≠ z_{2}

This shows that different element of C may have the same value in R.

Thus, f(z) is not one-one.

f is onto if every element of R is the f-image of some element of C.

We have, f(z) = |z|, ∀ z ∈ C and |z| ≥ 0

We observe that negative real numbers in R do not have their pre-images in C.

Thus, f is not onto.

Hence, f(z) is neither one-one nor onto.

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