Question

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

Answer 1

Given $f\left(z\right)=|z|$ for all $Z$ $ϵ$ $C$.

Let $z_1=x+iy$ and $z_2=x-iy$ $x,y$ $ϵ$ $R$

Step 1: Injective or One-one function :-

$|z_1|=\sqrt{x^2+y^2};|z_2|=\sqrt{x^2+y^2}|z_1|=|z_2|But{z}_1\ne z_2$

Hence, $f\left(z\right)$ is not a one-one function.

Step 2: Surjective or Onto function :-

Let $y$ $ϵ$ $R$

Let $y=-\sqrt{2}$

There does not exist any no. $z$ in $C$ such that

$|z|=-\sqrt{2}$

Therefore, f is not onto.

Hence, f is neither one-one nor onto.