Is it a coincidence that the jacobian matrix of differentiable complex functions is also the matrix isomorphic to complex numbers?

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.)

• 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. – John Hughes Sep 6 '15 at 23:40
• Cool. I can't believe I never thought of it that way. – Matthew11 Sep 6 '15 at 23:43
• Hey, until you asked, I hadn't thought of it that way either. Thanks! – John Hughes Sep 6 '15 at 23:45

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 $\mathbf J=M(J)$.