The Set of Complex Numbers as an Algebraic Field

# The Set of Complex Numbers as an Algebraic Field

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

Theorem 1: The set of complex numbers $\mathbb{C}$ is a field with the binary operations of addition, $+$, and multiplication, $\cdot$. |

- 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. $\blacksquare$