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I know that in general it is impossible to fully visualize functions and their derivatives when mapping $f:\mathbb{C} \to \mathbb{C}$, because doing so requires being able to "see" in $4$ dimensions. However, I've decided to smuggle in a function which should only require $3$ dimensions to visualize.

Let $f:\mathbb{C} \to \mathbb{C}$, $f(z) = xy + 0i$ for $z = x+yi$, and let $f'(w) = \lim_{z \to w} \dfrac {f(z) - f(w)}{z-w}$. Is there a nice visualization for the derivative?

Graph of $f$:

enter image description here

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    Your function $f$ is not differentiable in the sense of complex analysis (the limit you call $f'(w)$ is not well-defined). (Indeed, your $f$ is real-valued and the only real-valued functions of a complex variable that are complex differentiable are constant.) – 2017-02-20
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    @RobArthan Ah yes I remember my professor saying something like that in class, but I didn't really understand it; $f$ being constant would mean that $\forall z_1, z_2 \in \mathbb{C} f(z_1) = f(z_2)$. But I can find many complex numbers $z_1, z_2$ such that $f(z_1) \not = f(z_2)$. – 2017-02-20
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    There is no contradiction, because your $f$ is not complex differentiable: the limit you call $f'(w)$ is not well-defined. E.g., take $w = 1 + i$ and compare what happens when $z$ approaches $w$ horizontally ($z = 1+ t + i$) with what happens when $z$ approaches $w$ vertically ($z = 1 + (1 + t)i$). – 2017-02-20
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    @RobArthan Ah I think I understand. It's not that my $f$ is a constant, it's that my $f$ is not differentiable, right? – 2017-02-20
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    That's exactly right! – 2017-02-20
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    @RobArthan Thank you for your help! – 2017-02-20
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    A real nice book for visualising complex functions: [Visual complex functions by E. Wegert](http://www.visual.wegert.com/) – 2017-02-20

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