# The Field of Complex Numbers

We will now verify that the set of complex numbers $\mathbb{C}$ forms a field under the operations of addition and multiplication defined on complex numbers.

Let $z_1, z_2, z_3 \in \mathbb{C}$ such that $z_1 = a_1 + b_1i$, $z_2 = a_2 + b_2i$, and $z_3 = a_3 + b_3i$.

**FA1:**$z_1 + z_2 = (a_1 + a_2) + (b_1 + b_2)i$, and so $(z_1 + z_2) \in \mathbb{C}$.

**FA2:**$z_1 + z_2 = (a_1 + b_1i) + (a_2 + b_2i) = (a_1 + a_2) + (b_1 + b_2)i = (a_2 + a_1) + (b_2 + b_1)i = (a_2 + b_2i) + (a_1 + b_1i) = z_2 + z_1$.

**FA3:**$z_1 + (z_2 + z_3) = a_1 + (a_2 + a_3) + (b_1 + [b_2 + b_3])i = (a_1 + a_2) + a_3 + ([b_1 + b_2] + b_3)i = (z_1 + z_2) + z_3$.

**FA4:**The additive identity is $0 = 0 + 0i$. That is $z_1 + 0 = (a_1 + b_1i) + (0 + 0i) = (a_1 + 0) + (b_1 + 0)i = a_1 + b_1i = z_1$.

**FA5:**The additive inverse for $z_1$ is $-z_1 = -a – b_i$. That is $z_1 + (-z_1) = (a_1 + b_1i) + (-a_1 – b_1i) = (a_1 – a_1) + (b_1 – b_1)i = 0 + 0i = 0$.

**FM1:**$z_1 \cdot z_2 = (a_1 + b_1i)(a_2 + b_2i) = a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2 = (a_1a_2 – b_1b_2) + (a_1b_2 + b_1a_2)i$ and so $(z_1 \cdot z_2) \in \mathbb{C}$.

**FM2:**$z_1 \cdot z_2 = (a_1a_2 – b_1b_2) + (a_1b_2 + b_1a_2)i = (a_2a_1 – b_2b_1) + (a_2b_1 + b_2a_1)i = z_2 \cdot z_1$.

**FM3:**

(1)

\begin{align} z_1 \cdot (z_2 \cdot z_3) \\ = (a_1 + b_1i) [ (a_2a_3 – b_2b_3) + (a_2b_3 + b_2a_3)i ] \\ = (a_1a_2a_3 – a_1b_2b_3) + (a_1a_2b_3 + a_1b_2a_3)i + (b_1a_2a_3 – b_1b_2b_3)i + (b_1a_2b_3 + b_1b_2a_3)i^2 \\ \quad = [(a_1a_2a_3 – a_1b_2b_3) – (b_1a_2b_3 + b_1b_2a_3)] + [(a_1a_2b_3 + a_1b_2a_3) + (b_1a_2a_3 – b_1b_2b_3)]i \\ = [ (a_1a_2 – b_1b_2) + (a_1b_2 + b_1a_2)i ] (a_3 + b_3i) \\ = (z_1 \cdot z_2) \cdot z_3 \end{align}

**FM4:**The multiplicative identity is $1 = 1 + 0i$. That is $z_1 \cdot 1 = (a_1 + b_1)(1 + 0i) = (a_1 + a_1 \cdot 0i) + (b_1 + b_1 \cdot 0)i = a_1 + b_1i = z_1$.

**FM5:**The multiplicative inverse for $z_1$ where $z_1 \neq 0$ is $z_1^{-1} = \frac{a_1}{a_1^2 + b_1^2} + \frac{a_1 -b_1i}{a_1^2 + b_1^2}$. That is:

(2)

\begin{align} \quad z_1 \cdot z_1^{-1} = (a_1 + b_1i) \cdot \left (\frac{a_1 -b_1i}{a_1^2 + b_1^2} \right) = \frac{(a_1 + b_1i)(a_1 – b_1i)}{a_1^2 + b_1^2} = \frac{a_1^2 – a_1b_1i + a_1b_1i – b_1i^2}{a_1^2 + b_1^2} = \frac{a_1^2 + b_1^2}{a_1^2 + b_1^2} = 1 \end{align}

**D1:**$z_1(z_2 + z_3) = (a_1 + b_1i)([a_2 + a_3] + [b_2 + b_3]i) = a_1[a_2 + a_3] + a_1[b_2 + b_3]i + b_1[a_2 + a_3]i + b_1[b_2 + b_3]i^3$- $= [ (a_1a_2 – b_1b_2) + (a_1b_2 + b_1a_2)i ] + [ (a_1a_3 – b_1b_3) + (a_1b_3 + b_1a_3)i ] = z_1 \cdot z_2 + z_1 \cdot z_3$.

Since the set of complex numbers $\mathbb{C}$ satisfy all eleven axioms under the operations of addition and multiplication, it follows that $\mathbb{C}$ is a field.

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