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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.

An entire function is a holomorphic function whose domain of a function is the whole complex plane. Its derivative exists at every $z\in \mathbb{C}$. Every polynomial, the complex exponential function $e^z$, the product $z{e}^z$, and the complex sine function $\sin z$ are entire functions. The Taylor series of an entire function about any finite centre converges throughout the complex plane.

The complex exponential function is a particularly revealing entire function. Writing $z=x+iy$, Euler's formula givesThus the complex exponential function never takes the value zero. Conversely, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages under the complex exponential function areHere $\ln |w|$ is the real natural logarithm, and $k$ ranges over the integers. Substitution gives $e^{z_k}=|w|e^{i\theta }=w$; equality of moduli and arguments shows these are all the solutions. The complex exponential function therefore has image exactly $\mathbb{C}\setminus \{0\}$, with infinitely many preimages for each nonzero value.

In the pictured vertical strip $-1\le x\le 1$, the real part sets the modulus between $e^{-1}$ and $e$, while the imaginary part sets the argument modulo $2\pi$. The two drawn bands each make one full turn around the same annulus; their corresponding points differ by $2\pi i$ and have identical images. Allowing all real $x$ gives every positive modulus, so the whole complex plane covers every nonzero output.

An entire function has exceptionally strong global constraints. The Liouville theorem says that a bounded function which is also an entire function must be a constant function. Accordingly, the nonconstant entire functions above are unbounded somewhere in the complex plane, even when their restrictions to the real axis are bounded: the complex sine function satisfies $\sin (iy)=i\sinh y$, growing through the hyperbolic sine. The Little Picard theorem adds a different constraint: an unbounded nonconstant entire function still cannot leave two distinct complex values out of its image. These results make entire functions useful when global bounds or restrictions on the image can establish rigidity without solving for the function explicitly.

In the opening of his 1880 memoir, Émile Picard credits Karl Weierstrass with the terminology of entire functions and describes them by everywhere-convergent power series. ^[ref]

If an entire function $g$ satisfies $\mathrm{Re}(cg)\ge b$ for constants $c\ne 0$ and real $b$, then $e^{-cg}$ is entire and has modulus at most $e^{-b}$. The Liouville theorem makes it constant, and differentiating gives $g^{\prime }=0$ since the exponential never vanishes. The conclusion also applies after rotating and translating any containing half-plane; separate bounds on the real and imaginary parts are unnecessary.

Let $f$ be a holomorphic function on $0<|z-a|<R$, and suppose $a$ is an isolated essential singularity. There is a set $E\subseteq \mathbb{C}$ with at most one element such that, for every $w\notin E$ and every $0<r<R$, the equation $f(z)=w$ has infinitely many distinct solutions with $0<|z-a|<r$. Thus each nonexceptional value has preimages accumulating at $a$ in every punctured neighbourhood. The same possible exception works for all these neighbourhoods; it need not be a completely omitted value, since it could be attained only finitely often near $a$. ^[ref]

The basic example is $f(z)=e^{1/z}$ at $a=0$. It is a holomorphic function for $z\ne 0$, and its Laurent serieshas infinitely many nonzero negative-power terms, making zero an essential singularity. Its sole omitted value is zero. For the value one, its preimages are exactlyThey satisfy $e^{1/z_k}=1$ and $|z_k|=1/(2\pi |k|)\to 0$. For any $r>0$, every integer $k$ with $|k|>1/(2\pi r)$ therefore supplies a distinct solution inside $0<|z|<r$.

The diagram displays only finitely many preimages; the formula and its limit, rather than the finite picture, establish that infinitely many occur in every punctured neighbourhood. More generally, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages arewith any zero denominator excluded. Such an exclusion occurs only for $w=1$, with exactly one excluded integer $k$ for the chosen argument. The remaining preimages are distinct and tend to zero as $|k|\to \infty$.

The isolated essential singularity hypothesis is crucial. The function $f(z)=z^{-1}$ has a pole at zero; on $0<|z|<r$ its image is only $\{w:|w|>1/r\}$. It omits every value in the closed disc $|w|\le 1/r$, and each attained value has just one preimage. A pole therefore has entirely different local behaviour from an essential singularity.

The Casorati-Weierstrass theorem says that the image near an essential singularity is a dense subset of the complex plane. The Great Picard theorem replaces arbitrary approximation by exact attainment, and then repeats that attainment infinitely often. This is useful when a local restriction on the image can rule out an essential singularity and reduce the possibilities to a pole or a removable singularity. For a meromorphic function viewed as a holomorphic map into the Riemann sphere, the corresponding statement has at most two exceptional sphere values, including infinity in the count; it applies when the map has an isolated essential singularity, even if poles accumulate there. Picard's 1880 memoir proves the local result in section 15, treating the cases with and without accumulating poles separately. ^[ref]

The Great Picard theorem also explains the Little Picard theorem. For a nonconstant polynomial, every value is attained by the fundamental theorem of algebra. For an entire function which is not a polynomial, the function $g(\zeta )=f(1/\zeta )$ has an essential singularity at zero: a removable singularity would force $f$ to be constant by the Liouville theorem, and a pole would force $f$ to be a polynomial. Applying the Great Picard theorem to $g$ leaves at most one finite omitted value for $f$.

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 $e^{-f}$.

The rational function $F(z)=z^2/(1-z{)}^2$ is holomorphic on the unit disc and maps it onto the entire complex plane. For any $w\in \mathbb{C}$, choose a square root $r$ with $\mathrm{Re}r\ge 0$ and put $z=r/(1+r)$. The identity $|1+r{|}^2-|r{|}^2=1+2\mathrm{Re}r>0$ gives $|z|<1$, while $F(z)=r^2=w$. It is not injective: the two choices $r=\pm i$ give distinct points in the disc with image $-1$.

A holomorphic bijection in this direction is impossible: its inverse would be a bounded entire function, contradicting the Liouville theorem. Thus a surjection onto the complex plane can exist even when a conformal equivalence cannot.

Suppose $D\subseteq \mathbb{C}$ is open and admits an injective holomorphic map $\Phi :D\to \mathbb{C}$ whose image is bounded. Every entire function $f:\mathbb{C}\to D$ is constant: the function composition $\Phi \circ f$ is a bounded entire function, hence constant by Liouville theorem, and injectivity of $\Phi$ then forces $f$ to be constant. Surjectivity of $\Phi$ onto a disk is unnecessary. For example $z\mapsto z/(1-z)$ is a conformal map from $\mathbb{C}\setminus [0,1]$ onto $(\mathbb{C}\setminus [0,\infty ))\setminus \{-1\}$. A square-root branch with argument in $(0,2\pi )$ followed by the Cayley transform between the half-plane and disk maps this to the punctured unit disk. It therefore supplies a bounded injective holomorphic coordinate on the interval complement, giving the same rigidity even though that domain is not simply connected.

Every nonconstant entire function takes every complex value with at most one exception. Equivalently, its image is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a constant function has only one value. The Little Picard theorem guarantees existence of preimages, without requiring infinitely many preimages for each attained value. ^[ref]

For a nonconstant polynomial $p$, the fundamental theorem of algebra applied to $p(z)-w$ supplies a preimage for every $w\in \mathbb{C}$. The complex exponential function has precisely one omitted value, zero, as the explicit formula for its preimages shows. In contrast, the complex sine function has no omitted value. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the quadratic equationIts roots have product $-1$, so they are nonzero. Every such root has a preimage under the complex exponential function, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the periodicity of the complex sine.

The force of the Little Picard theorem is that it applies to every nonconstant entire function, without an explicit formula. The Liouville theorem already gives a dense image of a nonconstant entire function, but a dense subset could still leave out many individual points. The Little Picard theorem limits these missing points to one. In particular, establishing two omitted values proves that an entire function is constant.

For a nonconstant meromorphic function on the complex plane, the corresponding bound is two omitted values in the Riemann sphere $\mathbb{C}\cup \{\infty \}$, where poles take the value infinity. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ for $k\in \mathbb{Z}$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an entire function, infinity is already an omitted value in the Riemann sphere, leaving at most one additional finite omitted value.

Proof outline: use the version of Montel's theorem for two fixed finite omitted values. ^[ref] It makes the rescaled holomorphic functions $g_n(\zeta )=f(a+n\zeta )$ a normal family on the unit disc. The fixed value $g_n(0)=f(a)$ gives pointwise anchoring of a holomorphic normal family, hence a subsequence with a finite holomorphic function as its locally uniform convergence limit. The Cauchy integral formula bounds its derivatives at zero, whereas $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.

Assume a regular invertible normalized Riemann-Hilbert problem solution $\mu =I+{\mu }_1/k+{\mu }_2/k^2+\cdots$ with jump tending to the identity and conjugated by $e^{-i(kx+2k^{2}t){\sigma }_3}$. Residuals of its two differential equations have no jump and vanish at infinity, so Liouville theorem gives a Lax pair. Coefficient comparison gives the displayed reconstruction and $A=2Q$; compatibility yields $i{Q}_t-Q_{xx}{\sigma }_3+2Q^3{\sigma }_3=0$. A jump matrix tending to zero instead is inconsistent with the normalization. Recovering the initial jump from a prescribed initial potential still requires direct scattering.

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.

If an entire function $f=u+iv$ has $uv$ bounded above, then $f$ is constant. Indeed $F=\exp (-i{f}^2)$ is entire and $|F|=\exp (2uv)$ is bounded, so the Liouville theorem makes $F$ constant. Differentiating gives $f{f}^{\prime }=0$. The derivative of $f^2$ is therefore zero, and continuity on the connected complex plane makes $f$ itself constant. No separate boundedness of $u$ or $v$ is required.

If a holomorphic bijection existed, its inverse would be an entire function taking values in the unit disc. Liouville theorem would make that inverse constant, contradicting bijectivity. Thus no holomorphic bijection exists.

A continuous bijection does exist, even a homeomorphism:Both maps preserve the argument and change the radius by the mutually inverse strictly increasing functions $r\mapsto r/(1-r)$ and $s\mapsto s/(1+s)$. The displayed formulas also prove continuity at zero and everywhere else.

If a holomorphic bijection existed, its inverse would be an entire function taking values in the unit disc. Liouville theorem would make that inverse constant, contradicting bijectivity. Thus no holomorphic bijection exists.

A continuous bijection does exist, even a homeomorphism:Both maps preserve the argument and change the radius by the mutually inverse strictly increasing functions $r\mapsto r/(1-r)$ and $s\mapsto s/(1+s)$. The displayed formulas also prove continuity at zero and everywhere else.

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.

Set $F(z)=\exp [-(1-i)g(z)]$. It is entire, and $\mathrm{Re}[(1-i)g]=u+v$ makes $|F|=e^{-(u+v)}\le 1$. By the just-proved Liouville theorem, $F$ is constant. Differentiation gives $0=F^{\prime }=-(1-i)g^{\prime }F$. The exponential never vanishes, so $g^{\prime }=0$ everywhere and $g$ is constant. This is the entire function confined to a half-plane is constant principle.

The Liouville theorem states that every bounded entire function is constant. Let $|f(z)|\le M$ and choose distinct $a,b$. For $R>\max (|a|,|b|)$, partial fractions and the Cauchy integral formula givewith the circle traversed anticlockwise. On that circle,The displayed difference is independent of $R$, so $f(a)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant. This proves the theorem with the requested two-pole contour rather than assuming it.