$f: D(\subset \mathbb C)\to \mathbb C$ is differentiable & real-valued ($D$ being a domain) $\Rightarrow$ $f$ is constant in $D$. Also the usual metrics on $D$ & $\mathbb R^2$ are identical. Can't we say from the above two observations that for any $f$: An open ball in $\mathbb R^2 \to \mathbb R$, the differentiability of $f$ in $\mathbb R^2$ $\Rightarrow$ $f$ is constant?
The Differentiability of $f$ in $\mathbb R^2$
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0@JavierBadia Won't the same metrics produce the same analysis? – 2012-12-10
1 Answers
The point is, as mentioned in @Javier's comment, that differentiability in $\mathbb C$, means more than differentiability in $\mathbb R^2$. The former one means, that there is an $a \in \mathbb C$, such that $ f(x+ h) = f(x) + ah + o(h), \qquad h \to 0 \tag 1$ for each $x \in D$ and the latter that there is a linear $A \colon \mathbb R^2 \to \mathbb R^2$ such that $ f(x +h) = f(x) + Ah + o(h), \qquad h \to 0 \tag 2$ Rewriting the complex product in $(1)$ as a matrix product like in $(2)$, we see that $(1)$ implies that $A$ is of the special form $ A = \begin{pmatrix} \Re a & \Im a \\ -\Im a & \Re a\end{pmatrix} $ As there are more real $2\times 2$-matrices as these, there are linear maps on $\mathbb R^2$ which aren't $\mathbb C$-differentiable, for example $x\mapsto \overline x$, repesented by $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$.
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2It's worth noting that requiring $A$ (that is, $Df$) to be in that form is equivalent to the Cauchy-Riemann equations. – 2012-12-10