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I can follow and understand the algebra in the proof that a $\mathbb{R}$ differentiable function is $\mathbb{C}$ differentaible if and only if the partial$\left(\frac{\partial f}{\partial x}\right)$ -$\frac{1}{i}$ $\left(\frac{\partial f}{\partial y}\right)$ $ = 0$. But I can't picture this. Treating points in the complex plane like vectors in $\mathbb{R^2}$, if a function is differentiable in $\mathbb{R^2}$, why, geometrically, is it not always holomorphic?

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up vote 7 down vote accepted

What does "differentiable" mean geometrically? Locally approximationable by a linear map, right? So if a function $f \colon \mathbb R^2 \to \mathbb R^2$ is differentiable at $z \in \mathbb R^2$, we have that $$ f(x + h) = f(x) + Ah + o(h), \qquad h \to 0 $$ for a linear $A \colon \mathbb R^2 \to \mathbb R^2$. Being holomorphic means, that locally, $f\colon \mathbb C \to \mathbb C$ can be approximated by a complex linear map $\mathbb C \to \mathbb C$, that is written as $$ f(x + h) = f(x) + a\cdot h + o(h), \qquad h \to 0 $$ with $a \in \mathbb C$. Now ask yourself which linear maps $\mathbb R^2 \to \mathbb R^2$ arise from complex multiplication? Multiplication preserves angles at 0, for example, general linear maps needn't do that. Writing $a = \Re a + i\Im a$, and $h = \Re h + i \Im h$, we have that \[ a \cdot h = (\Re a \Re h - \Im a\Im h) + i (\Im a \Re h + \Re a \Im h) \] that is complex linear map are represented by real $2 \times 2$-matrices of the form \[ \begin{pmatrix} \Re a & -\Im a\\ \Im a & \Re a \end{pmatrix} \] But not all linear maps (which can arise as derivatives of a $\mathbb R$-differentiable function) look like this.

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+1 Nice explanation. Conjugation is an example, since $C(x+iy) = x-iy$, as a operator on $\mathbb{R}^2$ it is linear and has the form $C=\pmatrix{ 1 & 0 \\ 0 & -1}$ which cannot be expressed as multiplication by a complex number since the diagonal elements are not equal. – copper.hat Nov 8 '12 at 17:13

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