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You can "represent" complex numbers with 2x2 matrices, with the isomorphism between the fields $(\Bbb C,+,\times)$ and $(\begin{pmatrix} a & -b \\ b & a \end{pmatrix},+,\times)$: $$f:a+bi\mapsto\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ with $a,b\in\Bbb R$. Additionally, a complex function can be represented by $f(x+iy) = u +iv$, or that $f$ is composed of the two functions $u(x,y)$ and $v(x,y)$. The differentiable function $f$ has the property that its jacobian matrix $\begin{pmatrix} \partial u/\partial x & \partial v/\partial x \\ \partial u/\partial y & \partial v/\partial y \end{pmatrix}$ must be of the form$$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ Is this just a coincidence, or can we generalize this? For example, a split complex number $a+jb$ is represented by $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$. Would this matrix also have to do with the partial derivatives of a split-complex function? (Although I have not a slightest idea what it might mean to take the derivative of such a thing.)

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    $\begingroup$ It's not a coincidence. To be differentiable, the derivative at each point must be ... a complex number, which (in the matrix form of things) must be a matrix of that form. $\endgroup$ – John Hughes Sep 6 '15 at 23:40
  • $\begingroup$ Cool. I can't believe I never thought of it that way. $\endgroup$ – Matthew11 Sep 6 '15 at 23:43
  • $\begingroup$ Hey, until you asked, I hadn't thought of it that way either. Thanks! $\endgroup$ – John Hughes Sep 6 '15 at 23:45
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To elaborate on John Hughes' comment, recall that a function $(u,v):\mathbb R^2\mapsto\mathbb R^2$ is differentiable at $\mathbf z\in\mathbb R^2$ if there exist a matrix $\mathbf J$ such that $$ \lim_{\|\mathbf h\|\to0}\frac{\left\| \pmatrix{u(\mathbf z+\mathbf h)\\ v(\mathbf z+\mathbf h)} -\pmatrix{u(\mathbf z)\\ v(\mathbf z)} -\mathbf J\mathbf h\right\|}{\|\mathbf h\|}=0.\tag{1} $$ And a function $f=u+iv:\mathbb C\to\mathbb C$ is differentiable at $z\in\mathbb C$ if there exists a complex number $J$ such that $$ \lim_{h\to0}\frac{f(z+h)-f(z)-Jh}{h}=0.\tag{2} $$ Now define $M(x+iy)=\pmatrix{x&-y\\ y&x}$ and $\mathbf e_1=\pmatrix{1\\ 0}$. Then $M(z_1z_2)\equiv M(z_1)M(z_2)$ and every vector $\pmatrix{p\\ q}\in\mathbb R^2$ can be written as $M(p+iq)\mathbf e_1$. So, if $f$ is differentiable at $z$ and we rewrite $(2)$ in the form of $(1)$, we must have $$ \mathbf J\mathbf h=M(Jh)\mathbf e_1=M(J)M(h)\mathbf e_1=M(J)\mathbf h. $$ for every $\mathbf h\in\mathbb R^2$. Consequently $\mathbf J=M(J)$.

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