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In my notes I am told to take this as true without proof, which is cool... for some.

$f^{-1}$ is holomorphic by a property of holomorphic functions.

To show that $f'/f $ is Holomorphic, I think I would have to show that the limit

$$\lim_{h \rightarrow 0}\frac{f'(x+h)f(x)-f'(x)f(x+h)}{hf(x)f(x+h)}$$

If I have multiplied it out correctly, exists. $\forall x \in R \subset \mathbb{C}$

Am I along the correct path?

  • You might do this. Buut you might also already know that the product of two holomorphic functions is holomorphic, and that the composition of holomorphic functions is holomorphic, and that $z\mapsto z^{-1}$ is holomorphic on $\Bbb C\setminus{0}$. Therefore $\frac{f'(z)}{f(z)}=f'(z)\cdot(f(z))^{-1}$ is holomorphic. – Hagen von Eitzen May 05 '16 at 15:35
  • How do you know that $f'$ is holomorphic ?? –  May 05 '16 at 15:37
  • @HMPARTICLE Isn't a holomorphic function infinitly differentiable? In fact, it's equal to its own (absolutely) convergent power series. Wouldn't the derivative of an absolutely convergent power series again be an absolutely convergent power series? I put question marks because I'm not sure. – Fly by Night May 05 '16 at 16:05
  • I am unsure. If that is the case then I have no issue. but yes to the convergent power series part –  May 05 '16 at 16:25
  • http://math.stackexchange.com/questions/640/why-are-differentiable-complex-functions-infinitely-differentiable –  May 05 '16 at 16:30
  • The title is wrong because it claims $f'/f$ is holomorphic and non-zero – zhw. May 05 '16 at 20:33

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