1. **Problem:** Prove that the set of complex numbers $\mathbb{C}$ is a field under the usual operations of addition and multiplication. 2. **Definition:** A field is a set equipped with two operations (addition and multiplication) satisfying certain properties: closure, associativity, commutativity, existence of identity elements, existence of inverses, and distributivity of multiplication over addition. 3. **Step 1: Closure** - For any $z_1 = a + bi$ and $z_2 = c + di$ in $\mathbb{C}$, their sum and product are also in $\mathbb{C}$: $$z_1 + z_2 = (a+c) + (b+d)i \in \mathbb{C}$$ $$z_1 \cdot z_2 = (ac - bd) + (ad + bc)i \in \mathbb{C}$$ 4. **Step 2: Associativity and Commutativity** - Addition and multiplication of complex numbers are associative and commutative because real number operations are associative and commutative. 5. **Step 3: Identity Elements** - Additive identity: $0 + 0i$ satisfies $z + 0 = z$. - Multiplicative identity: $1 + 0i$ satisfies $z \cdot 1 = z$. 6. **Step 4: Inverses** - Additive inverse: For $z = a + bi$, the inverse is $-a - bi$. - Multiplicative inverse: For $z \neq 0$, inverse is $$z^{-1} = \frac{a - bi}{a^2 + b^2}$$ which is in $\mathbb{C}$. 7. **Step 5: Distributivity** - Multiplication distributes over addition: $$z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3$$ 8. **Conclusion:** All field axioms hold for $\mathbb{C}$ under usual addition and multiplication, so $\mathbb{C}$ is a field. **Final answer:** The set of complex numbers $\mathbb{C}$ with usual addition and multiplication forms a field.