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A degree-zero complex analytic divisor $E$ on a compact Riemann surface is principal exactly when its Abel-Jacobi map of a compact Riemann surface is zero. Normalize a differential of the third kind with residues equal to the coefficients of $E$. The Riemann bilinear relations for a compact surface express its $b$-periods as $2\pi i$ times the Abel integrals. If those integrals are in the period lattice, subtract a suitable integral multiple of the normalized holomorphic differential forms; all periods then lie in $2\pi i\mathbb{Z}$, so exponentiation gives the required meromorphic function. Conversely, the logarithmic differential of a meromorphic function has integer residues and periods in $2\pi i\mathbb{Z}$, giving a zero Abel class by the same calculation.

Strict increase alone does not suffice in the preceding theorem. Let ${\sigma }_j=1-1/j$, $c_j=2^{-j}$ and $f(y)=\eta (y){\sum }_{j\ge 1}{2}^{-j}{y}^{1-1/j}$, where $\eta$ is continuous, equals one on $(0,1]$ and vanishes on $[2,\infty )$. After division by the next power, every finite remainder extends continuously to zero. Nevertheless the Mellin transform has genuine poles at $-{\sigma }_j$ accumulating at $-1$, so it cannot be a meromorphic function on the whole plane.

Every nonconstant meromorphic function on the complex plane has a simple island over at least one of any five Jordan domains on the Riemann sphere with disjoint closures. The Ahlfors second fundamental theorem, area comparison over target domains, and a length-area exhaustion of the complex plane prove this: without a simple island every counted island has degree at least two, forcing $3S\le (5/2)S+O(L)$.

A meromorphic function $f$ has an algebraic addition theorem if some nonzero polynomial $P\in \mathbb{C}[U,V,W]$ satisfies $P(f(z),f(w),f(z+w))=0$ wherever the three values are finite. For a polynomial $f$ of degree $d>0$, take roots $z_i$ of $f(z)-U$ and $w_j$ of $f(w)-V$; the product ${\prod }_{i,j}(W-f(z_i+w_j))$ has polynomial coefficients in $U,V$ by the Fundamental theorem of symmetric polynomials, is monic in $W$, and supplies an addition relation. An elliptic function also has such a relation, by finite algebraic extension of its function field and the Weierstrass elliptic function addition formula.

For a meromorphic function near $0<|z|\le 1$, and a target $a$ off its image of the unit circle, let $u_a=-\log k(f,a)$ for the chordal metric, $M_a(s)$ its normalized angular mean, and $I(a)=M_a^{\prime }(1)$. The weighted spherical area $T(r)=(4\pi {)}^{-1}{\int }_{r<|z|<1}\log (|z|/r)(f_{\mathrm{round}}^{\#}{)}^{2}dA$ satisfies the displayed formula, where $N$ counts $a$-points with multiplicity and weight $\log (|z|/r)$ and $\overline{m}=M_a(r)-M_a(1)$. Indeed $\Delta u_a=(f_{\mathrm{round}}^{\#}{)}^2/2-2\pi \sum {\mathrm{deg}}_z(f){\delta }_z$, and Green's second identity with the radial logarithmic weight proves the formula. The angular mean must include $1/(2\pi )$.

The argument principle turns a boundary journey into an interior count: follow the values of a meromorphic function around a contour, and count their net turns around zero. Every enclosed zero of a function contributes its multiplicity of a zero; every enclosed pole subtracts its order. There is no need to find each zero of a function separately.

Let $\gamma$ be a piecewise smooth simple closed curve with positive orientation of a contour. Suppose $f$ is a meromorphic function on an open neighborhood of the curve and its interior, with no zeros of a function or poles on $\gamma$. ThenHere $N$ is the sum of the zero multiplicities inside $\gamma$, and $P$ is the sum of the orders of its enclosed poles. The image contour $f\circ \gamma$ may cross itself or revisit a point. Its signed winding number, rather than the number of visually bounded regions, is the quantity that counts.^[ref]

For the polynomial $f(z)=z^3$ and the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, the image is $e^{3it}$. It goes around zero three times although its drawn outline is just one circle. The interior contains one distinct zero of a function, at zero, with multiplicity of a zero three: the count is three. Reversing the orientation of a contour reverses the winding number and the sign of the contour integral.

An argument must be followed continuously along the journey. Subtracting the principal arguments at the endpoints loses whole turns: a closed image contour has the same endpoint value even when its winding number is nonzero. Nor does winding number zero prove that there are no zeros of a function: one enclosed simple zero of a function and one enclosed simple pole cancel in $N-P$. Known poles must be added back to recover $N$.

The mechanism is short. Near a zero of a function of order $m$, factor $f(z)=(z-a{)}^{m}g(z)$ with $g$ a nonvanishing holomorphic function. Its logarithmic derivative is $f^{\prime }/f=m/(z-a)+g^{\prime }/g$, so its residue is $m$. A pole of order $m$ gives residue $-m$. The residue theorem adds those contributions. Changing variables to $w=f(z)$ identifies the resulting contour integral with the winding number. This is a local algebraic count joined to a global geometric count.^[ref]

The usefulness goes beyond explicit polynomials. Rouché's theorem keeps a count unchanged when a boundary perturbation cannot push the image contour through zero. The Nyquist stability criterion applies the same counting idea to poles governing a system under feedback control.^[ref]

A holomorphic function on the unit disc has bounded characteristic when ${\sup }_{r<1}\int {\log }^{+}|f(r\xi )|dm(\xi )<\infty$, where ${\log }^{+}t=\max (0,\log t)$. Thus it belongs to the Nevanlinna class. A general meromorphic function requires the additional pole-counting term in its characteristic; the displayed analytic definition has no such term.

Contour shifting deforms a complex integration path through a region where the integrand is meromorphic. The residue theorem says that the change equals $2\pi i$ times the sum of residues of the poles crossed.

A meromorphic one-form on a compact Riemann surface with only simple poles is a differential of the third kind. Prescribed residues can be realized exactly when their sum is zero: in the residue exact sequence for $K_C(S)$, the obstruction in $H^1(C,K_C)\cong \mathbb{C}$ is the sum of residues. Subtracting a holomorphic differential form normalizes its $a$-periods. Integer residues and periods in $2\pi i\mathbb{Z}$ make the exponential of its integral a single-valued meromorphic function with those orders of zeros and poles.

The quotient sheaf ${\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$ records the zero and pole orders of local nonzero meromorphic functions. Each of its global sections is naturally a divisor on a complex manifold on $X$.

The connecting homomorphism ofsends a divisor on a complex manifold $D$ to its holomorphic line bundle associated to a divisor $\mathcal{O}(D)$ in the Picard group. Its kernel consists exactly of the principal divisors of global nonzero meromorphic functions.

The Weierstrass elliptic function satisfies the displayed meromorphic function identity. To prove it, the Riemann-Hurwitz formula shows that its derivative has exactly three simple zeros modulo its period lattice, at the nonzero two-torsion points of a complex torus. At each such half-period $h$, evenness around $h$ gives ${\wp }^{\prime \prime }/{\wp }^{\prime }=1/(z-h)+O(z-h)$, so the right side has principal part $1/(4(z-h{)}^2)$, matching $\wp (2z)$. At zero, $\wp =z^{-2}+O(z^2)$ and ${\wp }^{\prime \prime }/{\wp }^{\prime }=-3/z+O(z^3)$ give the same principal part and zero constant term. The difference is a holomorphic elliptic function, hence constant by compactness, and its limit at zero is zero.

An elliptic function is a meromorphic function with two real-linearly independent periods ${\omega }_1,{\omega }_2$. A fundamental cell is a half-open parallelogramOpposite boundary integrals cancel. Consequently a nonconstant elliptic function has equally many zeros and poles in a cell, counted with multiplicity, and the sum of its pole residues in a cell is zero.

The Weierstrass elliptic function is the quotient map from the complex torus by $z\mapsto -z$. An even elliptic function descends through this map; its even Laurent series at each branch point makes the descended function meromorphic there. A meromorphic function on the Riemann sphere is rational, so $f=Q(\wp )$.

A meromorphic function on a punctured disk has finite pulled-back spherical area precisely when it extends to the Riemann sphere at the puncture. A removable singularity or pole gives a smooth spherical extension and finite area. At an essential singularity, the meromorphic version of the Great Picard theorem gives infinitely many preimages of all but at most two spherical targets. The area formula then makes the area infinite. The logarithmically weighted annular area divided by $\log (1/r)$ tends to the total unweighted area divided by $4\pi$, by averaging a nondecreasing area function.

If a transcendental meromorphic function had only finitely many simple preimages over each of five distinct finite values, their simple counting functions would be $O(\log r)$. The displayed multiplicity inequality and the Nevanlinna first main theorem would bound the truncated sum by $5T_f/2+O(\log r)$. The Nevanlinna second main theorem requires at least $3T_f$ up to its $o(T_f)$ error, a contradiction. Thus at least one value has infinitely many simple preimages; four other values can remain totally ramified.

A nonidentity holomorphic automorphism of a compact connected Riemann surface of genus $g$ has at most $2g+2$ fixed points. If there were more, choose $g+1$ distinct fixed points and let $D$ be their sum. The Riemann-Roch theorem gives $\ell (D)\ge 2$, so there is a nonconstant meromorphic function with at most simple poles at these points and no other poles. The function $h=f-f\circ \phi$ has at most $g+1$ poles and at least $n-g-1$ zeros at the remaining fixed points. Since a nonzero principal divisor has degree zero, $n>2g+2$ forces $h=0$. At an actual pole fixed by $\phi$, the reciprocal $1/f$ is a local coordinate because its zero is simple. The identity $f=f\circ \phi$ therefore forces $\phi$ to be the identity near that pole, and the identity theorem on a Riemann surface makes it the identity everywhere.

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

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 $f$.

The growth order of a meromorphic function is the upper power exponent of its Nevanlinna characteristic, with constants assigned order zero. It is different from the local signed order at a zero or pole. A Nevanlinna integrated counting function has an analogous growth order. The Nevanlinna first main theorem bounds every counting order by the function order.