Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
Very nice answer, +1. –  Eric Naslund Feb 8 '11 at 17:41
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.