Could we define multiplication of “complex numbers” in this way?

Question

If we define multiplication of complex numbers as follows: $$z_1 \cdot z_2=(x_1x_2+y_1y_2, x_1y_2+x_2y_1)$$ then it can be shown that it induces a group structure $(G, \cdot)$, because it has inverse elements: $$z^{-1}=\left(\frac{x}{x^2-y^2},\frac{y}{-x^2+y^2}\right)$$ and the unit element appears also: $(1,0)$. It is also commutative and satisfies all other conditions (I think so). So why not define multiplication of complex numbers like this? Does this definition lead into contradiction? Of course you have no definition whatsoever for $\sqrt{-1}$, because now $i^2=1$ implies $i=1$.

Answer 1

This is just the ring $\mathbb{R}[t]/(t^2-1)$ in disguise; the equivalence class of the polynomial $a+bt$ corresponds to the ordered pair $(a,b)$, as you can see: $$\begin{align*} (x_1+y_1t)(x_2+y_2t)&=x_1y_1+(x_1y_2+x_2y_1)t+(x_2y_2)t^2\\\\ &\equiv (x_1y_1+x_2y_2)+(x_1y_2+x_2y_1)t\mod (t^2-1). \end{align*}$$ There's nothing "wrong" with this ring, but I think it is fair to say that it is less useful than the complex numbers.

Answer 2

Apparently you have given a group structure to $\mathbb{R}^{2}$ (apart from the addition). This is not the complex numbers. The definitive property of the complex numbers is algebraic closure, which is missing here. The multiplication structure you gave does not even make it into a field.