Proof of Canonical Form and Arithmetic of Complex Numbers

Let $\mathbb{C}$ be the field of complex numbers with imaginary unit $i$, and write $0_{\mathbb{C}}$ and $1_{\mathbb{C}}$ for the additive and multiplicative identities of $\mathbb{C}$. References to conditions 1, 2 and 3 are to the three conditions of The Complex Numbers. We use the field axioms in $\mathbb{C}$ and in $\mathbb{R}$, the latter being an ordered field by The Real Numbers, together with the field identity $u\cdot(-v)=-(u\cdot v)$, which follows from the distributive law as in the proof of Existence and Uniqueness of the Square Root of a Sum of Two Squares.

Claim 1. In $\mathbb{R}$ we have $0+0=0$, and by condition 1 the same equation holds when the sum is formed in $\mathbb{C}$. Adding to both sides the additive inverse of $0$ in $\mathbb{C}$ gives $0=0_{\mathbb{C}}$. Likewise $1\cdot1=1$ holds in $\mathbb{R}$, hence in $\mathbb{C}$; since $1\neq0$ in $\mathbb{R}$ and $0=0_{\mathbb{C}}$, the element $1$ is nonzero in $\mathbb{C}$ and so has a multiplicative inverse there, and multiplying both sides of $1\cdot1=1$ by it gives $1=1_{\mathbb{C}}$. Now let $a\in\mathbb{R}$. The equation $a+(-a)=0$ holds in $\mathbb{R}$, hence in $\mathbb{C}$ by condition 1, and its right-hand side is $0_{\mathbb{C}}$; therefore the real number $-a$ is the additive inverse of $a$ in $\mathbb{C}$. If moreover $a\neq0$, then $a\cdot(1/a)=1=1_{\mathbb{C}}$ holds in $\mathbb{C}$ for the same reason, so the real number $1/a$ is the multiplicative inverse of $a$ in $\mathbb{C}$.

Claim 2. Suppose $i\in\mathbb{R}$. By condition 1 the product $i\cdot i$ formed in $\mathbb{C}$ is then the product formed in $\mathbb{R}$, and Existence and Uniqueness of the Square Root of a Sum of Two Squares, applied to the real numbers $i$ and $0$, gives $0\le i\cdot i+0\cdot0=i\cdot i$. On the other hand, condition 2 and claim 1 give $i\cdot i=-1_{\mathbb{C}}=-1$, so $0\le-1$ in $\mathbb{R}$. Applying Existence and Uniqueness of the Square Root of a Sum of Two Squares to the real numbers $1$ and $0$ gives $0\le1$, and adding $-1$ to both sides gives $-1\le0$. By antisymmetry of the total order of $\mathbb{R}$ we get $-1=0$, hence $1=0$, contradicting the requirement $1\neq0$ in Field. Therefore $i\notin\mathbb{R}$.

Claim 3. Existence of a representation is exactly condition 3. For uniqueness, let $a,b,c,d\in\mathbb{R}$ satisfy $a+bi=c+di$. Adding to both sides the additive inverses of $c$ and of $bi$ and using commutativity and associativity of addition together with the distributive law in the form $di+(-(bi))=(d+(-b))i$, we obtain

where by claim 1 and condition 1 the elements $a+(-c)$ and $d+(-b)$ are real numbers. Suppose $d+(-b)\neq0$. Multiplying both sides by the multiplicative inverse of $d+(-b)$ in $\mathbb{C}$, which by claim 1 is the real number $1/(d+(-b))$, gives

and by condition 1 the right-hand side is a product of real numbers formed in $\mathbb{R}$, hence real. This contradicts claim 2. Therefore $d+(-b)=0$, that is $d=b$; and then $a+bi=c+bi$ gives $a=c$ after adding the additive inverse of $bi$ to both sides.

Claim 4. Let $a,b,c,d\in\mathbb{R}$. By commutativity and associativity of addition and the distributive law in $\mathbb{C}$,

and by condition 1 the sums $a+c$ and $b+d$ are the ones formed in $\mathbb{R}$. For the product, the distributive law gives

and commutativity and associativity of multiplication, condition 2, and the identity $u\cdot(-v)=-(u\cdot v)$ give

Collecting terms and using the distributive law once more,

where by condition 1 all the operations on real numbers displayed are those of $\mathbb{R}$.