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Every holomorphic map $V_1/{\Lambda }_1\to V_2/{\Lambda }_2$ lifts to $z\mapsto Az+x$, where $A$ is a complex-linear map satisfying $A{\Lambda }_1\subseteq {\Lambda }_2$. Period differences of a lift lie in the discrete target Euclidean lattice, so its derivative is periodic. Boundedness on a compact fundamental domain and the Liouville theorem make this derivative constant. The linear map is unique, and the translation vector is determined modulo ${\Lambda }_2$.

Let ${\pi }_{\Lambda }:\mathbb{C}\to \mathbb{C}/\Lambda$ and ${\pi }_{{\Lambda }^{\prime }}:\mathbb{C}\to \mathbb{C}/{\Lambda }^{\prime }$ be the universal covering maps of two one-dimensional complex tori. Every holomorphic map $f:\mathbb{C}/\Lambda \to \mathbb{C}/{\Lambda }^{\prime }$ has a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ satisfying ${\pi }_{{\Lambda }^{\prime }}\circ F=f\circ {\pi }_{\Lambda }$, and every such lift is an affine functionwith $a\Lambda \subseteq {\Lambda }^{\prime }$. Indeed, for $\lambda \in \Lambda$, the continuous function $F(z+\lambda )-F(z)$ takes values in the discrete set ${\Lambda }^{\prime }$ and is therefore constant. Hence $F^{\prime }$ is $\Lambda$-periodic. It is bounded on a compact fundamental parallelogram and therefore on the complex plane, so the Liouville theorem makes $F^{\prime }$ constant. If $f$ preserves the identity and the lift is chosen with $F(0)=0$, then $b=0$ and the lift is a linear map.

The radial map $z\mapsto z/(1+|z|)$ is a homeomorphism from $\mathbb{C}$ to the unit disc, with inverse $w\mapsto w/(1-|w|)$. A conformal equivalence in that direction would be a nonconstant bounded entire function, contradicting the Liouville theorem.

Composing a hypothetical conformal equivalence $\mathbb{C}\to \mathbb{H}$ with a Cayley transform $\mathbb{H}\to \mathbb{D}$ would give a nonconstant bounded entire function, contradicting the Liouville theorem.

A bounded harmonic function on the complex plane is constant. More generally, an entire harmonic function bounded on either side is constant: write it as $\mathrm{Re}f$ by harmonic function as the real part of a holomorphic function, then apply the Liouville theorem to $e^f$ or $e^{-f}$.

If an elliptic function had no poles, it would be entire. It is bounded on the closure of a fundamental parallelogram, and periodicity then makes it bounded on the whole complex plane. The Liouville theorem would make it constant. Thus every nonconstant elliptic function has a pole.

Every element of a nonzero complex unital Banach algebra has a nonempty compact spectrum of an element. If the resolvent of an element existed everywhere, composing it with any bounded linear functional would give a bounded entire function tending to zero at infinity. The Liouville theorem and point separation by the Hahn-Banach theorem would force the resolvent to be zero, contradicting the inverse equation. Nonunital algebras are treated by unitization.