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What does it mean for a complex function to be 'differentiable in the sense of real analysis'.

I understand that a function if complex differentiable, if and only if it is 'differentiable in the sense of real analysis' and it satisfies the Cauchy Rienmann equations. But how can a complex number, not on the real line, be differentiable in this sense?

Thanks!

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    $\begingroup$ remember you can always write a complex valued function $f$ in the form $u+iv$, with $u,v$ real valued $\endgroup$ – Federica Maggioni Apr 29 '13 at 13:46
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You can interpret a function $f: \mathbb{C}\to\mathbb{C}$ not only as a function on the complex plane, but also as a function $f: \mathbb{R}^2\to\mathbb{R}^2$ by interpreting it as a mapping $$ (x,y) \to (\textrm{Re } f(x+iy),\textrm{Im } f(x+iy) =: (f_1(x,y), f_2(x,y)) \text{.} $$

You can then apply $\mathbb{R}^2$ differentiation theory to $f$, e.g. ask whether the partial derivates $$ \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_2}{\partial x},\frac{\partial f_2}{\partial y}, $$ exist, and whether they can be used to compute $$ \frac{df(x,y)}{d(x,y)} \text{.} $$ Note that the Cauchy-Rienmann equations are equivalent to $\frac{df(x,y)}{d(x,y)}$ being a scaled rotation.

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The complex numbers can be identified with the real plane ($\mathbb R^2$) in a canonical way: $x+iy$ is mapped to $(x,y)$. Using this identification, we can think of a complex function $f:\mathbb C\to \mathbb C$ as a real function $f:\mathbb R^2\to \mathbb R^2$, and apply the definition of derivative for functions of real variables.

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