[motivation]
### Inverting the Exponential
The complex exponential is the entire function
\begin{align*}
\exp : \mathbb{C} &\to \mathbb{C} \setminus \{0\} \\
w &\mapsto e^w.
\end{align*}
It is periodic with period $2\pi i$: for every $w \in \mathbb{C}$ and every $k \in \mathbb{Z}$, $\exp(w + 2\pi i k) = \exp(w)$. In particular, $\exp$ is not injective — each value in $\mathbb{C} \setminus \{0\}$ has infinitely many preimages. If $z = r e^{i\theta}$ with $r > 0$ and $\theta \in \mathbb{R}$, then every solution of $e^w = z$ has the form
\begin{align*}
w = \ln r + i(\theta + 2\pi k), \qquad k \in \mathbb{Z},
\end{align*}
where $\ln r$ denotes the ordinary real logarithm of the positive number $r$. The [set](/page/Set) of all "logarithms of $z$" is the infinite discrete set $\{\ln r + i(\theta + 2\pi k) : k \in \mathbb{Z}\}$ — a vertical arithmetic progression in the complex plane with common difference $2\pi i$.
To define a "logarithm function," we must select exactly one of these infinitely many values for each $z \in \mathbb{C} \setminus \{0\}$. The question is whether this selection can be made continuously — and, ideally, holomorphically.
### The [Continuity](/page/Continuity) Obstruction
Suppose, seeking a contradiction, that there exists a continuous function $L : \mathbb{C} \setminus \{0\} \to \mathbb{C}$ satisfying $\exp(L(z)) = z$ for all $z \neq 0$. Consider the unit circle parametrised by $\gamma(t) = e^{it}$ for $t \in [0, 2\pi]$. The composition $t \mapsto L(e^{it})$ is a continuous function with
\begin{align*}
\exp(L(e^{it})) = e^{it} \qquad \text{for all } t \in [0, 2\pi].
\end{align*}
Since $L(e^{it})$ is a continuous logarithm of $e^{it}$, it must have the form $L(e^{it}) = it + 2\pi i k(t)$ for some function $k : [0, 2\pi] \to \mathbb{Z}$. But $L$ is continuous and $k(t)$ is integer-valued and continuous, so $k$ is constant: $k(t) = k_0$ for all $t$. Evaluating at the endpoints:
\begin{align*}
L(e^{i \cdot 0}) &= L(1) = 0 + 2\pi i k_0 = 2\pi i k_0, \\
L(e^{i \cdot 2\pi}) &= L(1) = 2\pi i + 2\pi i k_0.
\end{align*}
The first equation gives $L(1) = 2\pi i k_0$ and the second gives $L(1) = 2\pi i(k_0 + 1)$. These are contradictory.
The [topological](/page/Topology) explanation is that $\mathbb{C} \setminus \{0\}$ has nontrivial fundamental group $\pi_1(\mathbb{C} \setminus \{0\}) \cong \mathbb{Z}$: there exist closed curves (such as the unit circle) that cannot be continuously contracted to a point. The logarithm "detects" these curves via the winding number: each trip around the origin shifts the imaginary part of $\log z$ by $2\pi$.
### The Resolution: Cutting the Domain
The obstruction above relies on the existence of closed curves encircling the origin. If we remove a curve connecting the origin to infinity — say, the closed ray $(-\infty, 0]$ — the resulting domain $\mathbb{C} \setminus (-\infty, 0]$ is simply connected: every closed curve in this domain can be continuously deformed to a point without leaving the domain. In particular, no closed curve in $\mathbb{C} \setminus (-\infty, 0]$ winds around the origin, so the obstruction vanishes.
On this simply connected domain, we can make a continuous — in fact holomorphic — selection of $\log z$ by restricting the argument to the interval $(-\pi, \pi)$. The removed ray $(-\infty, 0]$ is the *branch cut*, the origin is the *branch point*, and the resulting single-valued holomorphic function is a *branch* of the logarithm.
[/motivation]