# Complex differentiabilty

I know that:

1) A function $f:\mathbb{R}^2\to \mathbb{R}^2$, when differentiable at a point, has a $2\times 2$ matrix as a derivative, which is a linear transformation from $\mathbb{R}^2\to \mathbb{R}^2$ best approximating the function linearly in some neighbourhood.

2) There is a ring homomorphism $\mathbb{C} \to Mat_{2x2}(\mathbb{R})$ as $a+ib \longmapsto \left[\begin{array}{11}a & -b\\b & a \end{array}\right]$

3) For a function $f:\mathbb{C} \to \mathbb{C}$, I can define complex differentiabilty as the best $\mathbb{C}$-linear approximation of the function locally at a point, i.e., $f'(z_0):h \mapsto f'(z_0)h$

Now, I want to combine these three observations, so that the Cauchy-Riemann equation falls out by considering a complex differentiable function as a function from $\mathbb{R}^2\to \mathbb{R}^2$ and connect the jacobian with the $\mathbb{C}$-linear transformation via the homomorphism.

I am having trouble even formulating a proposition that I can prove. Do I define something called 'Complexfying an $\mathbb{R}^2$-operator'? Any help will be appreciated.

The upshot will be that I can then 'shift' the proofs of some of the basic results of holomorphic functions (such as the fact that if the partial derivatives of the co-ordinate functions exist and are continuous then the function will be holomorphic, etc) to that of multivariable calculus.

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Lemma. Let $T:\mathbb{C}\to\mathbb{C}$ be $\mathbb{R}$-linear (considering $\mathbb{C}$ as an $\mathbb{R}$-vector space just by restricting the scalar multiplication). Then the following are equivalent:

• (i) T is $\mathbb{C}$-linear
• (ii) There exists $\lambda\in\mathbb{C}$ such that $T(z)=\lambda z$ for all $z\in\mathbb{C}$ (i.e. T is multiplication by $\lambda$).
• (iii) The matrix of $T$ w.r.t. the standard $\mathbb{R}^2$-basis is $\left[\begin{array}{11}a & -b\\b & a \end{array}\right]$, where $a+bi=\lambda$.

Now suppose $f:\mathbb{C}\to\mathbb{C}$ is complex differentiable at $z_0$, i.e. $T:\mathbb{C}\to\mathbb{C}$, $z\mapsto f'(z_0)z$ is the $\mathbb{C}$-linear derivative. Then $T$ is also the best $\mathbb{R}$-linear approximation of $f:\mathbb{R}^2\to \mathbb{R}^2$: $\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}=\left|\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)\right|\to 0$.

Hence the matrix of $T$ is the jacobian of $f$. Recalling that the matrix entries of the jacobian are the partial derivatives, from the Lemma we get the Cauchy-Riemann equations.

Conversely, suppose $f$ is totally differentiable as map $f:\mathbb{R}^2\to \mathbb{R}^2$ and the Cauchy-Riemann equations hold. Then there is a unique $\mathbb{R}$-linear map $T:\mathbb{C}\to\mathbb{C}$ which is the best approximation, i.e.

$\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}\to 0$.

We know its matrix entries are the partial derivatives. Since the Cauch-Riemann equations hold, the Lemma says that $T$ is $\mathbb{C}$-linear, say $T(z)=\lambda z$. Note that

$\left|\frac{f(z)-f(z_0)}{z-z_0}-\lambda\right|=\frac{\left|f(z)-f(z_0)-c(z-z_0)\right|}{|z-z_0|}=\frac{|f(x)-f(z_0)-T(z-z_0)|}{|z-z_0|}\to 0$.

Hence $T$ is also the complex derivative.

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Very nice answer, +1. – Eric Naslund Feb 8 '11 at 17:41

Nice question. The definition of a differentiable function $f : \mathbb{C} \to \mathbb{C}$ carries over verbatim from multivariable calculus: it is a function such that we can write $f(x + h) = f(x) + df_x(h) + o(|h|)$ for any $x$ and sufficiently small $h$, where $x, h \in \mathbb{C}$ and $df_x(h)$ is a linear operator $\mathbb{C} \to \mathbb{C}$.

Now we identify $\mathbb{C}$ with $\mathbb{R}^2$. Then the "realification" of $df_x(h)$ is the Jacobian of $f$ as a differentiable function $\mathbb{R}^2 \to \mathbb{R}^2$, and it is given by multiplication by a complex number, so your observations make sense.

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