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!

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