Let C be the set of complex numbers. Prove that the mapping `F:C to R` given by `f(z)=|z|, AA z in C,` is neither one-one nor onto.

Explanation

To prove that the mapping \( f: C \to R \) given by \( f(z) = |z| \) for \( z \in C \) is neither one-one nor onto, we will follow these steps:

### Step 1: Understanding the Function
The function \( f(z) = |z| \) represents the modulus of a complex number \( z \). A complex number can be expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers. The modulus is defined as:
\[ |z| = \sqrt{a^2 + b^2} \]

### Step 2: Show that the Function is Not Onto
To show that \( f \) is not onto, we need to demonstrate that there are real numbers in the codomain \( R \) that are not images of any complex number under \( f \).

- The modulus \( |z| \) is always non-negative, meaning \( |z| \geq 0 \).
- Therefore, the range of \( f \) is \( [0, \infty) \), which includes all non-negative real numbers.
- However, the codomain is \( R \), which includes negative numbers as well.

Since there are negative numbers in \( R \) that cannot be obtained as \( |z| \) for any \( z \in C \), we conclude that \( f \) is not onto.

### Step 3: Show that the Function is Not One-One
To show that \( f \) is not one-one, we need to find at least two different complex numbers \( z_1 \) and \( z_2 \) such that \( f(z_1) = f(z_2) \).

- Consider the complex numbers \( z_1 = 1 + i \) and \( z_2 = 1 - i \).
- We calculate their moduli:
\[ f(z_1) = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
\[ f(z_2) = |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
- We see that \( f(z_1) = f(z_2) = \sqrt{2} \), but \( z_1 \neq z_2 \).

Since we have found two different inputs that produce the same output, we conclude that \( f \) is not one-one.

### Conclusion
Thus, we have shown that the mapping \( f: C \to R \) given by \( f(z) = |z| \) is neither one-one nor onto.

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