Let $C$ be the set of complex numbers.
Prove that the mapping $f:\mathcal{C} \to R$
given by $f(z) = |z|,\forall z \in C$, is
neither one-one nor onto.

The mapping $f:C \to R$

Given $f(z) = |z|,\forall z \in C$

$f(1) = |1| = 1$

$f( - 1) = | - 1| = 1$

$f(1) = f( - 1)$

But $1 \ne - 1$

So, $f(z)$ is not one-one. Also, $f(z)$ is not

onto as there is no pre-image for any negative

element of $R$ under the mapping $f(z)$.