# Can we prove the Cauchy-Riemann equations using the matrix form of a complex number?

The definition of being complex-differentiable at $z$ can be stated as the existence of $a\in\mathbb C$ such that:

$$f(z+h)-f(z)=ah+r(h)|h|$$

For all $h$, where $r(z)\to0$ as $z\to0$. Thinking of complex numbers as elements of $\mathbb R^2$ in the obvious way, this is asserting the existence of a differential for $f:\mathbb R^2\to\mathbb R^2$ at $z$, where that differential is the mapping $z\to az$, which is of course a linear function (its obvious $\mathbb R^2\to\mathbb R^2$ analogue is linear, I mean).

But if the mapping if of the form $z\to az$, this implies that its matrix must have a very specific form. The linear mapping $(\Re (z), \Im (z))\to (\Re (az), \Im (az))$ can be shown to always have a matrix in the canonical basis of the form $\bigl(\begin{smallmatrix} x&-y\\ y&x \end{smallmatrix} \bigr)$. But when we compare that to the Jacobian matrix of $f$ at $z$, we get the Cauchy-Riemann equations.

Can this be made rigorous? In particular, how could one use it to prove the sufficiency of the CR equations? It seems to me the above only proves necessity.

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I've added a more specific question at the end. – Jack M May 6 '14 at 17:16

Suppose that $f$ is differentiable in $z \in \mathbb{C}$. Then there exists $a \in \mathbb{C}$ such that

$$f(z+h)-f(z) = a \cdot h + |h| r(h) \tag{1}$$

and $|r(h)| \to 0$ as $h \to 0$. Using the natural mapping

$$\mathbb{C} \ni (x+\imath y) \mapsto \begin{pmatrix} x \\ y \end{pmatrix}$$

we can consider elements of $\mathbb{C}$ as elements of $\mathbb{R}^2$. In particular, the multiplication of two complex numbers $$a \cdot h = (a_1+\imath a_2) \cdot (h_1+\imath h_2) = (a_1 h_1-a_2 h_2) + \imath (a_1 h_2 + a_2 h_1)$$ corresponds to the matrix multiplication $$\begin{pmatrix} a_1 & -a_2 \\ a_2 & a_1 \end{pmatrix} \cdot \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} = \begin{pmatrix} a_1 h_1-a_2 h_2 \\ a_2 h_1+a_1 h_2 \end{pmatrix}.$$

If we rewrite $(1)$, then we get

$$\begin{pmatrix} f_1(z_1+h_2,z_2+h_2) \\ f_2(z_1+h_1,z_2+h_2) \end{pmatrix}-\begin{pmatrix} f_1(z_1,z_2) \\ f_2(z_1,z_2 \end{pmatrix} = \begin{pmatrix} a_1 & -a_2 \\ a_2 & a_1 \end{pmatrix} \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} + |h| \begin{pmatrix} r_1(h_1,h_2) \\ r_2(h_1,h_2) \end{pmatrix}. \tag{2}$$

Now it follows from the definition of differentiability (of $\mathbb{R}^2$-valued functions) that $(f_1,f_2)$ is differentiable in $(z_1,z_2)$ and its Jacobian matrix equals

$$\begin{pmatrix} a_1 & -a_2 \\ a_2 & a_1 \end{pmatrix}. \tag{3}$$

Now suppose that $(3)$ is satisfied, i.e. that $(f_1,f_2)$ is differentiable in $(z_1,z_2)$ with Jacobian matrix of the form $(3)$. Again, by the definition of differentiability, $(2)$ holds for functions $r_1,r_2: \mathbb{R}^2 \to \mathbb{R}^2$ satisfying $|r_j(h_1,h_2)| \to 0$ as $|(h_1,h_2)| \to 0$. Setting $r(h_1+\imath h_2) := r_1(h_1,h_2)+\imath r_2(h_1,h_2)$ and considering the elements of $\mathbb{R}^2$ as elements of $\mathbb{C}$ we find that $(2)$ implies

$$f(z+h)-f(z) = (a_1+\imath a_2) \cdot h + |h| r(h).$$

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