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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]

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 $f$.

Every holomorphic map $f:\mathbb{C}\to \mathbb{C}/\Lambda$ is constant or surjective. It lifts to an entire function $F:\mathbb{C}\to \mathbb{C}$. If $F$ is nonconstant, the Little Picard theorem says it omits at most one point, while every fiber of $\mathbb{C}\to \mathbb{C}/\Lambda$ is an infinite lattice coset; hence every point of the torus has a preimage under $f$.

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.

Write $Y={\mathbb{P}}^1\setminus \{0,1,\infty \}$. One concrete uniformization uses the image in ${\mathrm{PSL}}_2(\mathbb{R})$ of the level-two principal congruence subgroupIt acts on the complex upper half-plane by Möbius transformations. Integer matrix entries make the group discrete, and the action is properly discontinuous. There are no nonidentity elliptic stabilizers: an elliptic element would have integer even trace with absolute value less than two, hence trace zero. But then $d=-a$ gives $bc=-(a^2+1)\equiv 2(\mathrm{mod}4)$, contradicting the evenness of $b,c$. Thus the quotient is an ordinary Riemann surface, with no orbifold points.

A fundamental region is the ideal quadrilateralIts vertical sides are identified by $T(\tau )=\tau +2$, and its semicircular sides by $U(\tau )=\tau /(1-2\tau )$. The ideal vertices are $-1,0,1,\infty$; $-1$ and $1$ represent one puncture, while zero and infinity give two others. Filling these three punctures gives genus zero: the identified quadrilateral has three vertices, two paired edges and one face, so its compactification has Euler characteristic $3-2+1=2$. Its complex compactification is a sphere, and a Möbius transformation puts the punctures at $0,1,\infty$. HenceThis is the thrice-punctured sphere as a modular quotient. Equivalently, the uniformization theorem applied to its universal covering map produces the upper-half-plane model: the sphere cannot cover this noncompact surface, and an entire covering from $\mathbb{C}$ would contradict Little Picard theorem by omitting both zero and one.

The curvature-minus-one metric $d{s}_{\mathbb{H}}=|d\tau |/\mathrm{Im}\tau$ is invariant under every real Möbius transformation preserving $\mathbb{H}$. Indeed, for determinant-one coefficients,Thus it descends to the hyperbolic metric on $Y$. This defines its distance by the infimum of path lengths; a choice of local lift does not affect lengths.

For the Schwarz-Pick lemma, first prove the Schwarz lemma. If $h:\mathbb{D}\to \mathbb{D}$ is a holomorphic function and $h(0)=0$, the function $h(z)/z$ extends as a holomorphic function across zero. On $|z|=r$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ yield $|h(z)|\le |z|$ and $|h^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative makes $h(z)/z$ a constant of modulus one, again by the maximum modulus principle, so $h$ is a rotation.

For a general holomorphic function $F:\mathbb{D}\to \mathbb{D}$, conjugate by the automorphisms of the unit disk $A_a(z)=(z-a)/(1-\overline{a}z)$, putting $a$ and $F(a)$ at zero. Applying the proved Schwarz lemma givesThe disk hyperbolic metric is $2|dz|/(1-|z{|}^2)$, and its distance is twice the inverse hyperbolic tangent of the displayed pseudohyperbolic distance. Both forms of Schwarz-Pick lemma therefore express metric contraction. Equality in the derivative at one point makes $F$ a disk automorphism by the equality case already proved.

For $X=\mathbb{D}\setminus \{0\}$ and a holomorphic function $f:X\to Y$, compose with a universal covering map of $X$ and lift to a universal covering map of $Y$. The simply connected source of the lift is a disk or half-plane. Schwarz-Pick lemma there bounds the lifted metric derivative, and the covering maps are local isometries, soThe second inequality follows by applying the length inequality to every joining path and taking the infimum. The hyperbolic metric on the punctured disk is its own intrinsic metric,obtained from the covering $z=e^{i\tau }$. It is not merely the ordinary disk metric restricted to the punctured disk.

Yes, equality can occur for distinct points. The cyclic subgroup $⟨T⟩$ lies in $\Gamma (2)$. Its quotient is conformally the punctured disk, using $z=e^{\pi i\tau }$, because this identifies precisely $\tau \sim \tau +2$. The subgroup inclusion induces a covering map that is a holomorphic functionIt is a local isometry. Choose a sufficiently small geodesically convex ball around any target point that is evenly covered. Two distinct sufficiently close points in one lift of that ball are joined by the lift of its minimizing target geodesic. The lift has the same length, giving $d_X\le d_Y$ for this pair, while contraction gives the reverse inequality. Thus equality holds.

For context, if the lifted map is not an automorphism, the derivative inequality is strict at every point; integrating along a minimizing geodesic in the complete punctured-disk metric gives a strict distance inequality for every distinct pair. Equality forces the covering-map case, but even a covering map need not preserve all distances globally because shorter paths can pass through other sheets. This is the distinction captured by equality in hyperbolic Schwarz-Pick contraction.

The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere $\hat{\mathbb{C}}$, the complex plane $\mathbb{C}$, or the unit disc $\mathbb{D}$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\hat{\mathbb{C}}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The Riemann surfaces uniformized by the complex plane are

Let $U\subseteq \mathbb{C}$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb{C}\to U\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting two values, contrary to the Little Picard theorem. The plane domain with two omitted points is hyperbolic, so

Now put $S=R\setminus \{P_1,\dots ,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities arewhich give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives

For the complex torus $X=\mathbb{C}/\Lambda$, any analytic map $f:\mathbb{C}\to X$ lifts through the universal covering map to an entire function $F:\mathbb{C}\to \mathbb{C}$. If $F$ is constant, so is $f$. Otherwise Little Picard says that $F$ omits at most one complex number. For every $x\in X$, its lifts form an infinite coset $z+\Lambda$, so at least one lift is attained by $F$. Hence the holomorphic map from the complex plane to a one-dimensional complex torus satisfies

Finally, $\mathbb{C}$ and $\mathbb{D}$ are homeomorphic byThey are not conformally equivalent: a biholomorphism $\mathbb{C}\to \mathbb{D}$ would be a nonconstant bounded entire function, contradicting the Liouville theorem. This proves the claimed example of complex plane and unit disc are homeomorphic but not conformally equivalent.

The uniformization theorem states that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere ${\mathbb{C}}_{\infty }$, the complex plane $\mathbb{C}$, or the open unit disc $\mathbb{D}$. Equivalently, every connected Riemann surface is a quotient of one of these by a free properly discontinuous group of conformal automorphisms.

For ${\mathbb{C}}_{\infty }$, every nonidentity Möbius transformation has a fixed point. A free action is therefore trivial, so the only surface uniformized by the sphere is

Every conformal automorphism of $\mathbb{C}$ has the form $z\mapsto az+b$. If $a\ne 1$, it has a fixed point, so a group acting freely consists only of translations. A discrete subgroup of $(\mathbb{C},+)$ has rank zero, one, or two. The corresponding quotients arewhere $\Lambda$ is a rank-two lattice and $\mathbb{C}/\Lambda$ is a complex torus. This proves the completeness of the Riemann surfaces uniformized by the complex plane.

Now let $D\subseteq \mathbb{C}$ be a domain whose complement contains two distinct points. Its universal cover cannot be the sphere because no nonconstant holomorphic map from a compact Riemann surface has image in the noncompact proper domain $D$. If its universal cover were $\mathbb{C}$, the covering map followed by $D\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting the two chosen complementary points, contradicting the Little Picard theorem. The uniformization theorem therefore leaves onlywhich is the plane domain with two omitted points is hyperbolic result.

Finally let $j:{\mathbb{C}}^{\ast }\hookrightarrow R$ be a holomorphic embedding with $R$ compact. At each puncture zero and infinity, an essential singularity would, by the Great Picard theorem in a coordinate chart, force values to be taken repeatedly and contradict injectivity. Thus $j$ extends holomorphically toA generic point of the open image $j({\mathbb{C}}^{\ast })$ has exactly one preimage, so $J$ has degree one. A degree-one holomorphic map between compact Riemann surfaces is a biholomorphism; equivalently, the Riemann-Hurwitz formula rules out branching and gives the same conclusion. Therefore the compact Riemann surface containing an embedded punctured plane is

The Picard theorem has two forms concerning the image of a holomorphic function. The Little Picard theorem permits at most one omitted value for a nonconstant entire function. The Great Picard theorem is local: near an isolated essential singularity, every complex value except possibly one occurs infinitely often. The complex exponential function $e^z$ and the function $e^{1/z}$ show that the possible exception is necessary.

Émile Picard presented the research in 1879 and published proofs in his 1880 memoir. Its opening footnote lists communications to the Académie des Sciences on 28 April, 19 May, and 20 October 1879; sections 3 and 15 contain the global and local results respectively. ^[ref]

Every plane domain whose complement contains at least two points is uniformized by the unit disc. A plane universal cover would give a nonconstant entire function omitting two values, contrary to the Little Picard theorem.