TITLE: Show that there does not exist a holomorphic function $h(z)$ such that $\exp(h(z)) = z$ on the punctured plane without using complex integration?
QUESTION [7 upvotes]: We must show that there does not exist a holomorphic function $h(z)$ on the domain $\mathbb C - \{0 \}$ such that $\exp(h(z)) = z$ on the complex plane. Can we do this without using complex integration?
I know the proof of the fact that there exists no function whose exponential is the identity function $id(z) = z$ on the complex plane which uses complex analysis.
We proceed by assuming such a  holomorphic $h(z)$ exists. This means that $e^{h(z)} = z$. Differentiating, we get $1 = e^{h(z)} h'(z) = z h'(z)$. This means that $1 = z h'(z)$, or  $h'(z) = 1/z$.
Now we compute $\oint h'(z)$ around the countour $c(t) = e^{2 \pi i t}$ in two different ways:

$\oint h'(z) = h(1) - h(1) = 0$, by using FTC and that the countor starts and ends at the same point: $1$.
$\oint h'(z) = \oint 1/z = 2 \pi i$ by the Residue Theorem

Thus, we get $0 = 2 \pi i$ which is absurd.
However, this proof seems to rely on a lot of the machinery of complex integration to get things done. Is there no "simpler" proof? Can we show that to prove this fact, we somehow "need to" invoke facts about complex integrals?
Ideally, I would want an answer that only uses elementary properties of complex numbers, and properties of the complex exponential, and complex differentiation, but not complex integration.

REPLY [2 votes]: There cannot be a continuous function $h$ on the unit circle $|z|=1$ such that $e^{h(z)}=z$. If there were, then a connectedness argument could be applied to prove that $h(e^{i\theta})=i\theta+2n\pi i$ for all $0 \le \theta < 2\pi$ and for some fixed integer $n$, which would contradict the assumed continuity of $h$ at $z=1$.