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I was reading this question where it says that an analytic complex function $f:D \to \mathbb{C}$ are open mappings when $f′(z)$ is never zero. I´m not clear on two things, pardon me if they are obvious but I'm just starting with complex analysis. The first one is why does the total derivative of a function $f:D \to \mathbb{R}^2$, seen as $f(x,y)= (f_1(x,y),f_2(x,y))$, is

$\begin{pmatrix} \partial_x f_1 & -\partial_x f_2 \\ \partial_x f_2 & \partial_x f_1 \end{pmatrix}$?

The second is why does the inverse function theorem tells us that $f$ is an open mapping if we have that $f′(z)$ is never zero?

Thank you in advance for any help.

Ben-ZT
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1 Answers1

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That matrix form is a consequence of the complex derivative being multiplication by a complex number (i.e. a rotation and scaling). This restriction gives the Cauchy–Riemann equations, which in turn allow you to substitute to remove the $y$ derivatives.

The inverse function theorem's content in this case is that when $f'(z)$ is not zero, there is a neighbourhood $D$ of $z$ so that $f$ is a bijection $D \to f(D)$, and this inverse, $g$ say, has derivative $g'(w)=1/f'(z)$, where $w=f(z)$ (i.e. $g'(w)=1/f'(g(w))$). Thus $g$ is also complex-differentiable, and hence continuous.

But $f$ is an open mapping if and only if the images of open sets are open, which is the same as saying that the inverse images of open sets under its inverse $g$ are open: if we write $B=g^{-1}(A)$, we also have $B=f(A)$, so if $A$ is open, either $B$ is always open and both [$g$ is continuous] and [$f$ is an open map] are true, or there is a nonopen $B$ and neither statement is.

Chappers
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