%
\label{thm:complex-field-axioms}
$\mathbb{C}_\tau$ is a \textbf{commutative field}.
That is:
\begin{enumerate}
    \item \textbf{Additive group.}
          $(\mathbb{C}_\tau, +)$ is an abelian group
          with identity $0 = 0 + 0i$
          and additive inverse $-(a + bi) = (-a) + (-b)i$.
    \item \textbf{Multiplicative group.}
          $(\mathbb{C}_\tau \setminus \{0\}, \cdot)$ is an abelian group
          with identity $1 = 1 + 0i$.
    \item \textbf{Distributivity.}
          Multiplication distributes over addition.
\end{enumerate}
In particular, every nonzero $z = a + bi \in \mathbb{C}_\tau$
has a multiplicative inverse:
\[
    \boxed{%
    (a + bi)^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\, i
    = \frac{a - bi}{a^2 + b^2}.}
\]

[Proof sketch]
Commutativity, associativity, and distributivity
are verified by direct computation
using the operations of Definition~\ref{def:tau-complex-field}.
For the multiplicative inverse,
suppose $z = a + bi \neq 0$,
so that $a \neq 0$ or $b \neq 0$.
Then $a^2 + b^2 > 0$ in $\mathbb{R}_\tau$
(since $a^2 \geq 0$, $b^2 \geq 0$,
and at least one is strictly positive).
Define $w := (a - bi)/(a^2 + b^2)$. Then:
\begin{align*}
    z \cdot w
    &= (a + bi) \cdot \frac{a - bi}{a^2 + b^2} \\
    &= \frac{(a + bi)(a - bi)}{a^2 + b^2} \\
    &= \frac{a^2 - (bi)^2}{a^2 + b^2}
    = \frac{a^2 - b^2 i^2}{a^2 + b^2}
    = \frac{a^2 + b^2}{a^2 + b^2}
    = 1.
\end{align*}
This is the crucial step:
the denominator $a^2 + b^2$ is strictly positive
\emph{because} $i^2 = -1$ causes the cross terms to add
rather than cancel.
Compare with the split-complex case below,
where $j^2 = +1$ causes the cross terms to cancel,
producing zero divisors.