Given $f(z)=|z|\;for\; all\; Z \in c$
Let $z_1=x+iy\;and\;z_2=x-iy \qquad x,y \in R$
Step1: Injective or One-One function:
$|z_1|=\sqrt {x^2+y^2}\qquad;\qquad|z_2|=\sqrt {x^2+y^2}$
$|z_1|=|z_2|$
But $z_1 \neq z_1$
Hence $f(z)$ is not a one-one function
Step 2: Surjective or On-to function:
Let $y \in R$
Let $y=- \sqrt 2 $
There does not exists any no. z in c such that $|z|=-\sqrt 2$
Therefore f is not onto
Solution: Therefore f is neither one-one nor onto