Let `C` and `R` denote the set of all complex numbers and all real numbers respectively. Then show that `f: C->R` given by `f(z)=|z|` for all `z in C` is neither one-one nor onto.

Options

A. f is injective but not surjective
B. f is surjective but not injective
C. f is nither injective nor surjective
D. f is both injective and surjective

Answer

To show that the function \( f: \mathbb{C} \to \mathbb{R} \) given by \( f(z) = |z| \) for all \( z \in \mathbb{C} \) is neither one-one nor onto, we will analyze both properties step by step.

### Step 1: Show that \( f \) is not one-one (injective)

1. **Definition of One-One Function**: A function \( f \) is one-one if \( f(z_1) = f(z_2) \) implies \( z_1 = z_2 \) for any \( z_1, z_2 \in \mathbb{C} \).

2. **Choose Two Different Complex Numbers**: Let \( z_1 = 3 + 4i \) and \( z_2 = 3 - 4i \). Clearly, \( z_1 \neq z_2 \).

3. **Calculate the Modulus**:
- For \( z_1 \):
\[ f(z_1) = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- For \( z_2 \):
\[ f(z_2) = |3 - 4i| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

4. **Conclusion**: Since \( f(z_1) = f(z_2) = 5 \) but \( z_1 \neq z_2 \), we conclude that \( f \) is not one-one.

### Step 2: Show that \( f \) is not onto (surjective)

1. **Definition of Onto Function**: A function \( f \) is onto if for every \( y \in \mathbb{R} \), there exists some \( z \in \mathbb{C} \) such that \( f(z) = y \).

2. **Range of \( f \)**: The modulus \( |z| \) for any complex number \( z \) is defined as:
\[ |z| = \sqrt{a^2 + b^2} \quad \text{where } z = a + bi \]
This value is always non-negative, meaning \( |z| \geq 0 \).

3. **Co-domain of \( f \)**: The co-domain of \( f \) is \( \mathbb{R} \), which includes all real numbers, both positive and negative.

4. **Conclusion**: Since \( f(z) \) can never take negative values, there are real numbers (e.g., \( -1 \)) for which there is no \( z \in \mathbb{C} \) such that \( f(z) = -1 \). Therefore, \( f \) is not onto.

### Final Conclusion

Since the function \( f(z) = |z| \) is neither one-one nor onto, we conclude that \( f \) is not bijective.

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