# Cauchy-Riemann Equations + Locally Invertible implies non-zero derivative

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 $$f(x,y) = (x^3, y^3)$$.

• Do you mean $Df(a) \neq 0$? – HallaSurvivor Jan 29 at 5:43
• @HallaSurvivor Yup, edited. – eigenvalues_question Jan 29 at 6:05
• Your "counterexample" is not holomorphic. – timur Jan 29 at 6:11
• @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. – eigenvalues_question Jan 29 at 6:18
• By holomorphy I mean Cauchy-Riemann is not satisfied at 0 for your "counterexample." – timur Jan 29 at 8:01

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 $$Df(0,0)=0$$.