The Set of Complex Numbers is a Field

# The Set of Complex Numbers is a Field

Recall that the set of real numbers is a field. We will now verify that the set of complex numbers is also a field.

Theorem 1: The set of complex numbers $\mathbb{C}$ with the binary operations of complex number addition, $+$ defined for all $a + bi, c + di \in \mathbb{C}$ by $(a + bi) + (c + di) = (a + c) + (b + d)i$, and with complex number multiplication defined for all $a + bi, c + di \in \mathbb{C}$ by $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$, forms a Field. |

- To show that $\mathbb{C}$ is a field with the operations $+$ and $\cdot$ we must show that $\mathbb{C}$ satisfies all of the field axioms with these operations. Let $z = a + bi, w = c + di, v = e + fi \in \mathbb{C}$.

**Field Axiom 1 (Closure Under Addition):**

\begin{align} \quad z + w = (a + bi) + (c + di) = (a + c) + (b + d)i \in \mathbb{C} \end{align}

**Field Axiom 2 (Commutativity of Addition):**

\begin{align} \quad z + w = (a + bi) + (c + di) = (a + c) + (b + d)i = (c + a) + (d + b)i = (c + di) + (a + bi) = z + w \end{align}

**Field Axiom 3 (Associativity of Addition):**

\begin{align} \quad z + (w + v) &= [a + bi] + ([c + di] + [e + fi]) \\ &= [a + bi] + ((c + e) + (d + f)i) \\ &= (a + (c + e)) + (b + (d + f))i \\ &= ((a + c) + e) + ((b + d) + f)i \\ &= ((a + c) + (b + d)i) + [e + fi] \\ &= ([a + bi] + [c + di]) + [e + fi] \\ &= (z + w) + v \end{align}

**Field Axiom 4 (Existence of an Additive Identity):**

\begin{align} \quad z + 0 = (a + bi) + (0 + 0i) = (a + 0) + (b + 0)i = a + bi = 0 \end{align}

**Field Axiom 5 (Existence of Additive Inverses):**

\begin{align} \quad z + (-z) = (a + bi) + (-a - bi) = (a - a) + (b - b)i = 0 + 0i = 0 \end{align}

**Field Axiom 6 (Closure Under Multiplication):**

\begin{align} \quad zw = (ac - bd) + (ad + bc)i \in \mathbb{C} \end{align}

**Field Axiom 7 (Commutativity of Multiplication):**

\begin{align} \quad zw = (ac - bd) + (ad + bc)i = (ca - db) + (da + cb)i = wz \end{align}

**Field Axiom 8 (Associativity of Multiplication):**

\begin{align} \quad z(wv) &= z[(ce-df) + (cf+de)i] \\ &= [(ace)-(adf)]-[(bcf) - (bde)] + [((acf)+(ade)) + (bce) - (bdf)]i \end{align}

- And also:

\begin{align} \quad (zw)v &= [(ac - bd) + (ad + bc)i]v \\ &= [(ace) - (bde)] - [(adf) + (bcf)] + [((acf) - (bdf)) + ((ade) + (bce))]i \end{align}

- It's not hard to see that these two equations are equal, so $(zw)v = z(wv)$.

**Field Axiom 9 (Existence of a Multiplicative Identity):**

\begin{align} \quad z \cdot 1 = (a + bi) \cdot 1 = a + bi = z \end{align}

**Field Axiom 10 (Existence of Multiplicative Inverses for Nonzero Elements):**For any $z \in \mathbb{C}$, $n \neq 0$, let:

\begin{align} \quad z^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i \end{align}

- Then:

\begin{align} \quad zz^{-1} &= (a + bi) \frac{a - bi}{a^2 + b^2} \\ \quad &= \frac{(a + bi)(a - bi)}{a^2 + b^2} \\ \quad &= \frac{a^2 + b^2}{a^2 + b^2} \\ \quad &= 1 \end{align}

- Since $\mathbb{C}$ satisfies all of the field axioms with the operations $+$ and $\cdot$ we have that $(\mathbb{C}, +, \cdot)$ is a field. $\blacksquare$