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I know that if $U$ is an open subset of the complex plane $\mathbb{C}$ and $f:U\to \mathbb{C}$ is a holomorphic function and $f$ is one-to-one, then the derivative $f'(z)$ is different from zero for every $z\in U$.

But how to prove converse of this problem:

If $f'(z)\neq 0,  \;\forall z\in U$, then $f(z)$ is locally injective?

Bernard
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Pascal
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  • Have a look at the power series expansion of your function $f$. What does the condition $f'(z)\neq 0$ tell you? –  Jul 01 '18 at 19:42
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    I'm thinking about whether there's a simple way to do this without citing any substantial theorems of complex analysis. Let $w=f(z).$ Let $dw = f(a+dz) - f(z),$ where $dz$ is an infinitely small compex number. Then $dw = f'(a),dz.$ If $f'(a)\ne0,$ then the mapping $dz\mapsto f'(a), dz$ is clearly one-to-one. HOWEVER, I wonder whether, in order to show there's an open neighborhood of $a$ within which $f$ is one-to-one, we might need to know that $f'$ is continuous in that neighborhood, as does not always happen with everywhere differentiable functions on $\mathbb R. \qquad$ – Michael Hardy Jul 01 '18 at 19:53
  • @Michael Hardy But, we don't have mean value theorem for complex valued function. – Pascal Jul 01 '18 at 21:10
  • @Pascal : What did you have in mind doing with a mean value theorem? – Michael Hardy Jul 01 '18 at 21:25
  • You'll find a complete proof using only basic techniques from complex analysis in "Complex Analysis" by Gilman et. al. It's Proposition 6.21 on pages 110/111. (See here: https://books.google.de/books?id=3jPOVGOilgIC&pg=PA108&lpg=PA108&dq=holomorphic+function+locally+injective&source=bl&ots=y6RoOOwdCk&sig=H0cM-3CBPmVh1OK4bNWlaulmiPk&hl=de&sa=X&ved=0ahUKEwidiujO9__bAhWLmbQKHTFkB24Q6AEIfzAJ#v=onepage&q=holomorphic%20function%20locally%20injective&f=false). But I strongly recommend trying it yourself. –  Jul 02 '18 at 07:58

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