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Suppose $C^1$ $f:\mathbb{R}^2 \to \mathbb{R}^2$ satisfies the Cuachy-Riemann equations $\frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial y}, \frac{\partial f_1}{\partial y} = -\frac{\partial f_2}{\partial x}$ at some point $a \in \mathbb{R}^2$. Additionally, let $f$ be locally invertible. Is it true that $Df(a) \neq 0$? I have been asked to prove that in a homework, but I cannot figure out how to do it without further conditions (e.g. $f^{-1}$ is differentiable). Any tips would be greatly appreciated.

It seems like there are obvious counter examples as given, like $f(x,y) = (x^3, y^3)$.

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    $\begingroup$ Do you mean $Df(a) \neq 0$? $\endgroup$ – HallaSurvivor Jan 29 at 5:43
  • $\begingroup$ @HallaSurvivor Yup, edited. $\endgroup$ – eigenvalues_question Jan 29 at 6:05
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    $\begingroup$ Your "counterexample" is not holomorphic. $\endgroup$ – timur Jan 29 at 6:11
  • $\begingroup$ @timur This is for a real analysis class, so we have not discussed holomorphisms. I am pretty sure $f:\mathbb{R}^2 \to \mathbb{R}^2$ with $f(x,y) = (x^3, y^3)$ is $C^1$, although maybe I am missing something obvious. $\endgroup$ – eigenvalues_question Jan 29 at 6:18
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    $\begingroup$ By holomorphy I mean Cauchy-Riemann is not satisfied at 0 for your "counterexample." $\endgroup$ – timur Jan 29 at 8:01
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The function $$f:\quad{\mathbb R}^2\to{\mathbb R}^2,\qquad(x,y)\mapsto(x^3,y^3)$$ is $C^1$, and even globally invertible, if we define (as usual) $$\root 3\of t:={\rm sgn}(t)\root3\of{|t|}\qquad(t\in{\mathbb R})\ .$$ Furthermore we have $$f_{1.1}=f_{2.2}=0,\qquad f_{1.2}=-f_{2.1}=0$$ at $(0,0)$, and therefore $Df(0,0)=0$.

The claim in your homework (as it appears in the question) can therefore not be proven.

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  • $\begingroup$ That is what I was thinking as well. I'll have to ask the professor what he meant on the homework. Thanks for the help. $\endgroup$ – eigenvalues_question Jan 29 at 15:44
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    $\begingroup$ @MichaelBurr The question was very clear that the Cauchy-Riemann equations at a point. $\endgroup$ – eigenvalues_question Jan 29 at 15:44
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I think the professor in our Honors Analysis class stated that local invertibility assumes a smooth inverse which (x^3,y^3) does not have at (0,0).

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  • $\begingroup$ I don't see any reason why a local inverse must be smooth. A smooth function with an invertible derivative will have a smooth inverse, but that isn't what the question is asking. $\endgroup$ – eigenvalues_question Jan 29 at 15:40

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