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

Concept: Function Properties (One-One and Onto)

A function is one-one if distinct inputs give distinct outputs, and onto if every element of the codomain has a preimage.

Step 1 – Not one-one:

Take $z_1 = 1$ and $z_2 = -1$. Both are in $\mathbb{C}$ and $f(1) = |1| = 1$, $f(-1) = |-1| = 1$. Since $1 \neq -1$ but $f(1) = f(-1)$, the function is not injective.

Step 2 – Not onto: …

We need to check two properties: injectivity (one‑one) and surjectivity (onto).

The function is $f(z)=|z|$, the distance of $z$ from the origin in the complex plane.

A single real output corresponds to infinitely many complex inputs — that immediately suggests it cannot be one‑one.

And since a distance is never negative, the codomain $\mathbb{R}$ contains numbers that are never reached — so it cannot be onto.

Let’s verify each formally.

1. One‑one (injective)

A function is one‑one if $f(z_1)=f(z_2)$ implies $z_1=z_2$.

Take $z_1 = 1$ and $z_2 = -1$. Both are in $\mathbb{C}$.

So $f(1)=f(-1)$ but $1 \neq -1$.

That’s a direct counterexample. Hence $f$ is not one‑one.

2. Onto (surjective)

A function is onto if every element of the codomain $\mathbb{R}$ is the image of some $z \in \mathbb{C}$. …

Method: Disproving one-one and onto by counterexample

To show a function is neither injective nor surjective, one concrete counterexample for each suffices.

Steps

Step 1: Break one-one with two equal outputs

Find distinct inputs $x_1\ne x_2$ with $f(x_1)=f(x_2)$. For the modulus map, $z=1$ and $z=-1$ both give $|z|=1$, so it is not one-one.

Step 2: Break onto by exposing a missed codomain value

Identify the actual range and find a codomain element outside it. Since $|z|\ge 0$, no complex number maps to a negative real, so the map onto $\mathbb{R}$ is not surjective. …

Common Mistakes

Mistake 1: Thinking "onto" just means the range is non-empty

Why it's wrong: onto requires the range to equal the whole codomain; here the range $[0,\infty)$ misses negative reals. Correct approach: compare range against codomain and exhibit a missed value like $-5$.

Mistake 2: Trying to prove non-injectivity in general instead of one counterexample …