# A continuous inverse of the exponential function is holomorphic

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!

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.

• Very nice and elementary proof! – Giorgos Giapitzakis Jun 20 at 19:57

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, $$\oint_{|z|=1}\frac{\mathrm dz}z=2\pi i$$.

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