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Let $\Omega \subseteq \mathbb{C}$ be open and connected such that $0 \notin \Omega$. Let $f: \Omega \to \mathbb{C}$ be a continuous function such that $e^{f(z)} = z$ for all $z \in \Omega$. Prove that $f$ is holomorphic and that $f'(z) = \frac{1}{z}$.

After asserting the first part I guess the second part is pretty trivial by just taking the derivatives of both sides (since $e^{f(z)}$ is holomorphic if $f(z)$ is holomorphic).

There is also a second part asking if there exists a continuous function $f: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ such that $e^{f(z)} = z$ for all $z \in \mathbb{C} \setminus \{0\}$. The answer for this question shoul be no because in class we defined $\log(z)$ as the inverse of $e^z$ and showed that it is continuous in $\mathbb{C}\setminus (-\infty, 0]$ and discontinuous everywhere else.

Is what I said correct? How would I go about proving the first part? Thanks in advance!

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Let $z \in \Omega $ and $(z_n)$ be a sequence in $\Omega\setminus \{ z \}$ with $z_n \to z$. Then $f(z_n) \to f(z)$ because $f$ is continuous, and $f(z_n) \ne f(z)$ because $f$ is necessarily injective. It follows that $$ \frac{f(z_n)-f(z)}{z_n-z} = \left(\frac{e^{f(z_n)}-e^{f(z)}}{f(z_n)-f(z)}\right)^{-1} \to \left( e^{f(z)}\right)^{-1} = \frac 1z $$ because $\exp$ is differentiable at $f(z)$ with $\exp' = \exp$. This demonstrates that $f$ is complex differentiable with $f'(z) = 1/z$.

More generally: If $f: \Omega \to D$ is continuous at $z_0 \in \Omega$, $g:D \to \Omega$ is holomorphic with $g(f(z)) = z$ for all $z \in \Omega$, and $g'(f(z_0)) \ne 0$, then $f$ is complex differentiable at $z_0$ with $$ f'(z_0) = \lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} = \lim_{z\to z_0} \left(\frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}\right)^{-1} = \left( g'(f(z_0))\right)^{-1} \, . $$

In other words: If $g$ is holomorphic, $g'\ne 0$, the existence of an inverse function $f$ is given and $f$ is assumed to be continuous, then the holomorphy of $f$ follows directly from the definition of the complex derivative.

For the second part of your question, see e.g. No holomorphic logarithmn.

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  • $\begingroup$ Very nice and elementary proof! $\endgroup$ – Giorgos Giapitzakis Jun 20 at 19:57
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Since $(\forall z\in\Bbb C):\exp'(z)\ne0$, $\exp$ is locally invertible. So, take $z_0\in\Bbb C$. There is some neighborhood $N$ of $f(z_0)$ such that $\exp|_N$ is an holomorphic inverse $l$. Since $f$ is continuous at $z_0$, there is some neighborhood $W$ of $z_0$ such that $f(W)\subset N$. So, since$$(\forall z\in W):\exp(f(z))=z,$$you have$$(\forall z\in W):f(z)=l(z).$$Therefore, $f$ is differentiable at $z_0$. So, $f$ is holomorphic.

And if there was some function $f\colon\Bbb C\setminus\{0\}\longrightarrow$ such that $(\forall z\in\Bbb C\setminus\{0\}):e^{f(z)}=z$, then $f'(z)=\frac1z$, and so $f$ would be a primitive of $\frac1z$. But then $\oint_{|z|=1}\frac{\mathrm dz}z=0$. But, in fact, $\oint_{|z|=1}\frac{\mathrm dz}z=2\pi i$.

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  • $\begingroup$ Thanks for the answer. Is there a more elementary proof since I'm quite new to complex analysis? $\endgroup$ – Giorgos Giapitzakis Jun 20 at 19:46
  • $\begingroup$ If there is, I don't know it. $\endgroup$ – José Carlos Santos Jun 20 at 19:48

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