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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.

Every complex elliptic curve, as a complex Lie group, is the quotient of $\mathbb{C}$ by a rank-two lattice. The coordinate functions are built from the Weierstrass elliptic function and its derivative. Holomorphic group homomorphisms lift to scalar multiplication on $\mathbb{C}$, and their endomorphism condition is $aL\subseteq L$. This turns complex multiplication into a statement about lattice stabilizers.

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.

Every elliptic function has a unique expression $A(\wp )+B(\wp ){\wp }^{\prime }$ with rational functions $A,B$. Its even part is rational in the Weierstrass elliptic function; its odd part divided by ${\wp }^{\prime }$ is even and meromorphic, so is also rational in $\wp$. The local even Laurent series at a half-period and at the pole justify meromorphic descent through the double cover of the Riemann sphere.

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 $f=Q(\wp )$.

The Weierstrass elliptic function has degree two on its period torus. Its three distinct half-period values $e_j$ have double local preimages, and its pole is double, so these four values are totally ramified. This also follows from ${\wp }^{\prime 2}=4{\prod }_j(\wp -e_j)$ and the principal part at the pole. A Möbius transformation with pole away from these values moves all four to finite values without changing local degrees.

The three nonzero half-period classes make the odd derivative of the Weierstrass elliptic function vanish. Their values are distinct: two equal values would force at least four zeros for a function with only one double pole. Thus they are three distinct roots of $4X^3-g_{2}X-g_3$, whose monic discriminant gives the displayed factor. This lattice discriminant differs by $(2\pi {)}^{12}$ from the normalized modular discriminant for the lattice $\mathbb{Z}+\mathbb{Z}\tau$.

The given Weierstrass elliptic function is even, by replacing the lattice index $\omega$ with $-\omega$. Differentiation givesFor any $\lambda \in \Lambda$, reindexing the sum shows ${\wp }^{\prime }(z+\lambda )={\wp }^{\prime }(z)$. Thus $\wp (z+\lambda )-\wp (z)=c_{\lambda }$ is constant off the lattice. Replacing $z$ by $-z-\lambda$ and using evenness changes this difference to its negative, so $c_{\lambda }=-c_{\lambda }$ and $c_{\lambda }=0$. Hence $\wp$ is doubly periodic.

It therefore descends to a meromorphic function on $\mathbb{C}/\Lambda$, with a double pole at zero modulo the lattice and no others. A meromorphic function is a holomorphic map to the Riemann sphere, sending poles to infinity. Locally at zero, $1/\wp (z)=z^2+O(z^4)$ is holomorphic and vanishes there, so the sphere chart at infinity supplies the required extension. The map is nonconstant because it has a pole.

Set $g(w)=a\wp (w)+b{\wp }^{\prime }(w)+c$. The Weierstrass elliptic function is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne 0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.

The point $\Phi (0+\Lambda )=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne 0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its principal divisor on the torus isApplying the elliptic divisor-sum identity givesThis is the period lattice form of the chord-and-tangent group law on an elliptic cubic.

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ and $T=\mathbb{C}/\Lambda$. Its fundamental parallelogram with opposite sides identified is a torus, so $\chi (T)=0$. The Weierstrass elliptic function is even and doubly periodic. One way to verify the latter from its series is to differentiate: the locally normally convergent derivative series iswhich is absolutely convergent away from the lattice and periodic by reindexing. Thus $\wp (z+{\omega }_j)-\wp (z)$ is constant. Evaluating at $z=-{\omega }_j/2$ and using evenness shows that this constant is zero for each generating period.

Therefore $[z]\mapsto \wp (z)$ descends to a meromorphic function on $T$, with one double pole at $[0]$. Viewed as a map to the Riemann sphere, a pole is a holomorphic point in the target coordinate $1/w$. The map is consequently holomorphic everywhere and

Multiplication by two on the complex torus is a holomorphic unramified coveringIts kernel is $(\frac{1}{2}\Lambda )/\Lambda$, of order four; its derivative is the nonzero constant two. Therefore $\mathrm{deg}[2]=4$. Composition with the degree-two Weierstrass elliptic function givesIt is well-defined because changing $z$ by $\omega \in \Lambda$ changes $2z$ by the period $2\omega$.

Take a real modulus $0<k<1$ and complementary modulus $k^{\prime }=\sqrt{1-k^2}$. The Jacobi elliptic sine is constructed by inverting a Schwarz-Christoffel mapping. The main reason for the Schwarz-Christoffel derivative is local angle behavior: if a prevertex $a_j$ corresponds to a polygon angle $\pi {\alpha }_j$, straightening that corner gives $U(w)-U(a_j)\sim C_j(w-a_j{)}^{{\alpha }_j}$. Hence $U^{\prime }$ has exponent ${\alpha }_j-1$. Along each straight boundary side its argument is constant. Dividing $U^{\prime }$ by ${\prod }_j(w-a_j{)}^{{\alpha }_j-1}$ removes those angle changes; Schwarz reflection extends the quotient across the real boundary and the corner singularities. With all vertices finite, its behavior at infinity is also regular, because the sum of the derivative exponents is $-2$. It is therefore a nonzero constant on the sphere. This yieldsThis argument explains the exponents, the branch choices and the role of boundary reflection, rather than just writing down the formula.

For a rectangle, four angles are $\pi /2$, so choose prevertices $-1/k,-1,1,1/k$ and normalize the inverse map asUse the branch positive on $(-1,1)$. SetThe two middle corner values are $U(\pm 1)=\pm K$. The right side length isas the substitution $t=(1-k^{\prime 2}{s}^2{)}^{-1/2}$ shows. The selected branch makes the increment along that side $i{K}^{\prime }$. Thus the other corner values are $K+i{K}^{\prime }$ and $-K+i{K}^{\prime }$. The two boundary intervals through infinity have combined horizontal length $2K$; the substitution $t=1/(ks)$ gives length $K$ from $1/k$ to infinity. In particular $U(\infty )=i{K}^{\prime }$. As in the square mapping, the boundary goes once around a rectangle, and the argument principle gives a conformal bijectionIts inverse $s(z)=\mathrm{sn}(z,k)$ maps that rectangle onto the complex upper half-plane, and near zero has $s(0)=0$, $s^{\prime }(0)=1$. Differentiating the inverse relation givesThese identities continue meromorphically.

All four rectangle edges have real images, so Schwarz reflection extends $s$ across them. Reflection in the bottom edge gives $s(\bar{z})=\overline{s(z)}$. Reflection in the right edge, combined with the first reflection, gives $s(2K-z)=s(z)$. The even inverse-integral derivative gives $s(-z)=-s(z)$, and henceReflection in the top edge and then the bottom gives $s(z+2i{K}^{\prime })=s(z)$. The resulting period lattice of the Jacobi elliptic sine isTo see the poles, use $U^{\prime }(w)\sim -1/(k{w}^2)$ near the boundary point infinity. Then $U(w)-i{K}^{\prime }\sim 1/(kw)$, so the inverse has a simple pole at $i{K}^{\prime }$ with residue $1/k$. The real half-period shift gives another pole at $2K+i{K}^{\prime }$ with opposite residue. Reflection tiles the plane with copies of the rectangle, so these are exactly two simple poles per displayed fundamental cell. They also show there is no additional period: a period must permute these two pole classes; exchanging them would shift by $2K$ modulo the displayed lattice, which reverses the function's sign and residues rather than preserving it.

For any finite $w$, integrate $s^{\prime }(z)/(s(z)-w)$ around a fundamental parallelogram avoiding its zeros and poles. The opposite edges cancel by periodicity, so the number of zeros of $s-w$ equals the number of poles, namely two. Therefore every value has two inverse images modulo periods, counted with multiplicity. The value infinity likewise has the two simple poles as preimages. Usually the finite-value preimages are distinct; the involution $z\mapsto 2K-z$ interchanges them. At $w=\pm 1,\pm 1/k$, a critical point supplies one double inverse image instead. For example $s(K)=1$, $s^{\prime }(K)=0$, $s^{\prime \prime }(K)=-(1-k^2)\ne 0$. Thus the printed “exactly two” requires the standard multiplicity convention; it would be false if interpreted as two distinct solutions at these branch values of a holomorphic map.

Now choose $z_0=K$ and put $f(z)=s(z+K)$. Since $s(2K-z)=s(z)$, $f$ is even, elliptic for $\Lambda$, and has degree two. The Weierstrass elliptic function for the same lattice is even and has one double pole per cell, hence also degree two. Its generic fibers are precisely $\{z,-z\}$. Consequently an even $f$ is constant on those fibers and descends to a meromorphic function $R$ of $\wp$. At the four fixed classes of $z\mapsto -z$, the even local Laurent expansion makes this descent meromorphic in the squared local coordinate. A meromorphic function on the Riemann sphere is rational, so $f=R(\wp )$. Generic map degrees multiply:Thus $R$ has degree one and the degree-two even elliptic function conclusion isThe determinant condition follows because $f$ is nonconstant. This proves the existence requested, with the explicit allowable shift $z_0=K$.

A covering-surface formulation of the Ahlfors second fundamental theorem is as follows. For a regular finite bordered Riemann surface $X$ mapped holomorphically to a fixed metric bordered Riemann surface $Y$, let $S=A_f(X)/A(Y)$ be the average sheet number and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the Euler characteristic,where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_{D}L$ for each fixed regular target region $D$. These formulations and the sign convention are explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf .

The island version of the Ahlfors second fundamental theorem for a disk source, which is the useful form here, says that for $q\ge 3$ fixed Jordan domains $D_j$ with disjoint closures on the Riemann sphere,Here $n_j$ counts simply connected islands of a meromorphic function over $D_j$, once each without weighting by mapping degree, andAn island of a meromorphic function is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General Jordan domains can be enclosed in slightly larger regular Jordan domains still having disjoint closures. A simple island over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.

Suppose a nonconstant meromorphic function on the plane has no simple islands over five such domains. On every counted simply connected island its proper map has integer degree at least two. The area formula therefore givesTogether with area comparison and the Ahlfors second fundamental theorem, this yieldsThe area comparison can also be seen directly. The two-form $[1_D/A(D)-1/(4\pi )]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its Poisson equation: the gradient of the spherical Green function has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying Stokes theorem, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.

We now derive the needed length-area exhaustion of the complex plane. The Cauchy-Schwarz inequality on a circle givesIf $L(R)/S(R)\ge \epsilon >0$ for all sufficiently large $R$, then $S(R)>0$ andIntegration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to \infty$ with $L(R_k)/S(R_k)\to 0$. This contradicts $S/2\le CL$ and proves the Ahlfors five islands theorem: one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it.

Four domains do not suffice. Take the Weierstrass elliptic function $\wp$ for the square lattice $\mathbb{Z}+i\mathbb{Z}$. The equationshows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all poles are double. Equivalently, the degree-two map from the elliptic torus to the Riemann sphere has these four branch values, with all their preimages of local degree two. Choose four small disjoint Jordan domains, each containing one of these four totally ramified Weierstrass values. A degree-one island of a meromorphic function over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus $\wp$ provides four domains with no simple island.

Choose an oriented basis ${\omega }_1,{\omega }_2$ of the period lattice, so $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. The Weierstrass elliptic function is constructed by the normally convergent seriesOn compact sets away from the lattice the summands are $O(|\omega {|}^{-3})$, which is summable over a plane lattice. It is even, has double poles at the lattice with zero residues, and is periodic. To check periodicity without splitting divergent series, differentiate termwise: ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is invariant under lattice shifts. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evenness evaluated at $z=-{\omega }_j/2$ makes this constant zero.

The Weierstrass sigma function is the canonical productIts logarithmic tails are likewise $O(|\omega {|}^{-3})$, so it defines an entire function with simple zeros exactly at the lattice and ${\sigma }^{\prime }(0)=1$. Pairing opposite lattice factors makes it odd. Its logarithmic derivative is the Weierstrass zeta functionThus ${\eta }_j=\zeta (z+{\omega }_j)-\zeta (z)$ is constant. Integrating this identity and evaluating the sigma ratio at $z=-{\omega }_j/2$ givesIntegration of zeta around a fundamental cell, containing one pole of residue one, gives the Legendre relation for Weierstrass quasi-periods ${\eta }_1{\omega }_2-{\eta }_2{\omega }_1=2\pi i$. These formulas explain why sigma itself is not an elliptic function, but balanced products of translates can be.

For completeness, the basic algebraic property of wp is $({\wp }^{\prime }{)}^2=4{\wp }^3-g_2\wp -g_3$, where $g_2=60{\sum }_{\omega \ne 0}{\omega }^{-4}$ and $g_3=140{\sum }_{\omega \ne 0}{\omega }^{-6}$. The identity follows by canceling the principal parts using the Laurent expansion: the difference is an entire elliptic function, hence constant, and its constant term is zero. The discriminant is nonzero; equivalently the three half-period values are distinct. These properties lead to the cubic embedding in 2(ii).

Now let $F$ be a nonzero meromorphic function on the torus and lift it to a periodic meromorphic function on the plane. Its logarithmic derivative $g=F^{\prime }/F$ is periodic. Choose a fundamental parallelogram with no zero or pole on its boundary. The residue theorem applied to $g$ gives equality of the total zero and pole multiplicities. Applied to $zg$, it gives the difference of the sums of the zero and pole lifts. Pairing opposite edges explicitly yieldsEach integral of $F^{\prime }/F$ on the right is $2\pi i$ times an integer, since its endpoint values of $F$ coincide and the image path avoids zero. Therefore the zero sum and pole sum agree in $\mathbb{C}/\Lambda$.

Conversely, suppose the point sums agree modulo the lattice. Change one chosen lift of a point by a lattice element so the two sums agree exactly in $\mathbb{C}$; this changes no point of the torus. Thenhas the prescribed divisor. Its multiplier under either basis period is $\exp ({\eta }_j(\sum Q_i-\sum P_i))=1$, so it descends to the torus. Repetitions contribute their appropriate multiplicities. This proves the principality criterion on a complex elliptic curve:There is an implicit qualification in the wording about separate zero and pole divisors: their supports must be disjoint. If a point occurs on both lists, cancel common multiplicities first when prescribing the divisor difference. A function cannot simultaneously have a zero and a pole at the same point; for example identical one-point lists satisfy the sum condition but do not specify separate zero and pole divisors. For $n=0$, any nonzero constant supplies the empty divisor.

The Weierstrass elliptic function and its derivative give the maponto the Weierstrass equation of an elliptic curve $Y^{2}Z=4X^3-g_{2}X{Z}^2-g_3{Z}^3$. Near zero, multiplying these coordinates by $z^3$ gives $[z+O(z^5):-2+O(z^4):z^3]$, so the map extends holomorphically and is an immersion at the origin.

The meromorphic map $\wp :E\to {\mathbb{P}}^1$ has degree two, since its only pole on the torus has order two. Its evenness implies $\wp (u)=\wp (v)$ exactly when $u\equiv v$ or $u\equiv -v$. Oddness of ${\wp }^{\prime }$ distinguishes these two alternatives unless they are the same point class. At each nonzero half-period, ${\wp }^{\prime }=0$. These three points exhaust its zeros, since its only pole has order three. The three half-period values are distinct: equality of two of them would give two distinct zeros each of order at least two for $\wp -e$, contradicting its total pole order two. They are roots of $4X^3-g_{2}X-g_3$, so the polynomial has distinct roots and the cubic is smooth. At a half-period $h$ with value $e$, ${\wp }^{\prime \prime }(h)=2{\prod }_{e^{\prime }\ne e}(e-e^{\prime })\ne 0$. Elsewhere ${\wp }^{\prime }$ is nonzero. The coordinate pair therefore has nonzero derivative at every finite point, and the map is injective. Compactness and the smooth cubic equation then identify it as an embedding onto the cubic.

The hyperplane section at infinity has divisor $3[O]$. Thus the intersection divisor $D_{\ell }$ of any line with this cubic is linearly equivalent to $3[O]$. Applying the principality criterion on a complex elliptic curve to $D_{\ell }-3[O]$ shows that its three points, counted with intersection multiplicity, have sum zero in $E$.

Conversely, if the degree-three divisor $D=[u]+[v]+[w]$ has point sum zero, Question 1 gives a meromorphic function with divisor $D-3[O]$. A meromorphic function with no poles away from $O$ and pole order at most three lies inTo prove this description, subtract multiples of ${\wp }^{\prime }$ and $\wp$ to cancel the cubic and quadratic Laurent principal parts. The remaining possible simple-pole coefficient is a residue, which must be zero because it is the only pole on a compact torus. The remaining holomorphic elliptic function is constant. Thus the constructed function is a linear form in the projective coordinates, and its line has intersection divisor exactly $D$. This proves the chord-and-tangent collinearity criterion is $u+v+w=0$ in $E$, with multiplicities understood.

For three pairwise distinct point classes away from $O$, ordinary collinearity is equivalent to linear dependence of the coordinate columns, soThe original PDF has this three-column determinant; the TeX's extra copied column is a transcription defect.

There is also a genuine implicit distinctness condition in the printed determinant assertion. If $u=v={\omega }_1/10$ and $w={\omega }_1/5$, all entries are finite and the determinant is zero because two columns coincide, but $u+v+w=2{\omega }_1/5\notin \Lambda$. Thus the unrestricted literal equivalence is false. To specify repeated intersections, use the confluent elliptic collinearity determinant: for a double point at $u$ replace the second column by $c^{\prime }(u)$, where $c(z)=(1,\wp (z),{\wp }^{\prime }(z){)}^t$; for a triple point use $c(u),c^{\prime }(u),c^{\prime \prime }(u)$. These impose tangency or inflection, and recover the correct conditions $2u+w\in \Lambda$ or $3u\in \Lambda$. If a point is $O$, use the homogeneous coordinates or a local homogeneous lift, since the displayed affine entries have poles there.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe summands on a compact set away from the lattice are $O(|\omega {|}^{-3})$, and that lattice sum converges. Thus the series and its termwise derivative converge normally. The function is even, and ${\wp }^{\prime }=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is odd and periodic by reindexing its absolutely convergent series. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evaluation at $-{\omega }_j/2$ and evenness make this constant zero for each basis period. The only poles of wp are double lattice poles, with principal part $(z-\omega {)}^{-2}$.

Let $G_r={\sum }_{\omega \ne 0}{\omega }^{-r}$ for the convergent weights used here. Expanding around zero givesThereforePut $g_2=60G_4$ and $g_3=140G_6$. The elliptic function $({\wp }^{\prime }{)}^2-4{\wp }^3+g_2\wp +g_3$ has a removable singularity at zero and hence at every lattice point. It is entire and periodic, thus bounded on the plane and constant by the Liouville theorem. Its zero constant term makes that constant zero. This proves the Weierstrass elliptic differential equationThe three classes ${\omega }_i/2$, with ${\omega }_3=-{\omega }_1-{\omega }_2$, are the three distinct nonzero two-torsion classes. Oddness and periodicity give ${\wp }^{\prime }({\omega }_i/2)=-{\wp }^{\prime }({\omega }_i/2)$, so they are zeros of the derivative. Their half-period values of the Weierstrass elliptic function $e_i$ are therefore roots of $4X^3-g_{2}X-g_3$.

They are distinct. If $e_i=e_j=e$ with $i\ne j$, the elliptic function $\wp -e$ would have a zero of multiplicity at least two at each of those distinct point classes, since its derivative vanishes there. It has only one pole on the torus, of order two. The argument principle on a fundamental cell equates its total zero and pole orders, contradicting at least four zeros. Thus $e_1,e_2,e_3$ are distinct, and the cubic factors asFor the monic cubic $X^3-(g_2/4)X-g_3/4$, the polynomial discriminant is $(g_2^3-27g_3^2)/16={\prod }_{i<j}(e_i-e_j{)}^2$. Hence the nonvanishing discriminant of a complex lattice isHere $\Delta (\Lambda )$ is the lattice discriminant, not the normalized q-series of Question 1. For $\Lambda =\mathbb{Z}+\mathbb{Z}\tau$, the normalizations give $g_2=(4{\pi }^4/3)E_4$ and $g_3=(8{\pi }^6/27)E_6$, so $\Delta (\Lambda )=(2\pi {)}^{12}\Delta (\tau )$.

If $\varphi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\varphi (x_0)>0$ and put $I_{+}=\{x\ge x_0:{\varphi }^{\prime }(x)>\varphi (x{)}^2\}$. Monotonicity implies $\varphi (x)>0$ thereafter. On this set, $1<{\varphi }^{\prime }/{\varphi }^2$, whenceAdding the finite initial interval $[0,x_0]$ provesThis derivative growth outside a finite-measure set controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=r{T}_s^{\prime }(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,Their logarithms are consequently $O(\log r+\log (T_s+1))$. Such derivative/area bounds in estimates from the Poisson-Jensen formula give the Nevanlinna logarithmic derivative lemma, $m(r,f^{\prime }/f)=O({\log }^{+}{T}_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.

We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental meromorphic function has $T_f(r)/\log r\to \infty$. Here is a justification of the latter fact. If there are infinitely many poles, fixing any arbitrarily large finite pole count gives $N(r;\infty )\ge A\log r-O_A(1)$. If there are only finitely many poles, multiply by a polynomial $p$ removing them to obtain a transcendental entire function $h=pf$. Its Taylor series has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the subharmonic function Poisson estimate gives $\log M_h(r/2)\le 3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.

Let $N_{\mathrm{simple}}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge 2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give ${\sum }_{j=1}^5\overline{N}(r;a_j)\le (5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le \sum \overline{N}+o(T_f)$, a contradiction. This proves the stronger result five values force infinitely many simple preimages:In particular more than one simple zero is necessary, but more than one target with simple zeros is not necessary.

For the last distinction, take a nonsingular Weierstrass elliptic function scaled so that ${\wp }^{\prime 2}=4\wp (\wp -1)(\wp +1)$; it has branch values $-1,0,1,\infty$. These are the four totally ramified Weierstrass values: finite branch preimages have local degree two, and all poles are double. SetThe first four targets are the images of those branch values under the Möbius transformation. Their preimages remain double; at a lattice pole, for example, $f(z)\sim (z-z_{\ast }{)}^2$. The last target corresponds to $\wp =3$, where ${\wp }^{\prime 2}=96\ne 0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant elliptic function is transcendental because it has infinitely many poles. Exactly one of the five selected finite targets therefore has simple preimages.

The local estimate gives $\wp (z)=z^{-2}+O(z)$ at zero and, by periodicity, at every lattice point. If two candidates existed, their difference would extend to an entire doubly periodic function: the principal parts cancel at every possible pole. It is bounded on a fundamental parallelogram, hence on the plane, so Liouville theorem makes it constant. The local $O(z)$ difference tends to zero, making the constant zero. Thus the function is unique.

The function $z\mapsto \wp (-z)$ has the same periodicity, poles and local estimate. Uniqueness therefore proves $\wp (-z)=\wp (z)$. Its Laurent expansion is consequentlywith no constant term by the given estimate. Periodicity and evenness imply $\wp (1/2+t)=\wp (1/2-t)$, so ${\wp }^{\prime }(1/2)=0$. The elliptic zero-pole counting theorem says a nonconstant elliptic function has, in one period cell, equally many zeros and poles counted with multiplicity. Applied to $\wp (z)-\wp (1/2)$, whose total pole order is two, it gives precisely two zeros counted with multiplicity. The zero at $1/2$ already has multiplicity at least two. It therefore has exactly two, provingFinally ${\wp }^{\prime \prime }-6{\wp }^2$ is periodic, and the Laurent expansion shows cancellation of all possible pole terms: ${\wp }^{\prime \prime }=6z^{-4}+2c_2+O(z^2)$ and $6{\wp }^2=6z^{-4}+12c_2+O(z^2)$. The difference extends to an entire bounded function and is constant. HenceEvery property of the Weierstrass elliptic function used here has been obtained from the stated local and periodic properties, together with the explicitly stated standard elliptic-function counting theorem.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe subtraction is essential: the separate inverse-square lattice sum need not converge. Let $r$ be the distance to the nearest nonzero lattice point. For $|z|<r$, the power series identity $(1-w{)}^{-2}={\sum }_{m\ge 0}(m+1)w^m$ givesThe stipulated convergence allows the sums to be interchanged on smaller disks. Pairing $\omega$ with $-\omega$ cancels odd powers. In terms of the lattice Eisenstein series, this proves the Laurent coefficients of the Weierstrass elliptic function:

For completeness, the function is even and is a periodic function: differentiating gives ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$, which is invariant under translation by a lattice generator. Thus $\wp (z+\omega )-\wp (z)$ is constant. For a primitive generator, evaluating at $z=-\omega /2$ and using evenness makes this constant zero. Two generators establish lattice periodicity. Hence any polynomial in $\wp ,{\wp }^{\prime }$ is an elliptic function.

Set $c_1=3G_4$ and $c_2=5G_6$. The first terms are $\wp =z^{-2}+c_1{z}^2+c_2{z}^4+O(z^6)$. Squaring its derivative and cubing the function giveConsequently $H={\wp }^{\prime 2}-4{\wp }^3+20c_1\wp +28c_2$ has no pole at zero and has value zero there. Lattice periodicity removes the poles at every lattice point; there are no others. Thus $H$ is an entire function and an elliptic function, bounded on a closed fundamental parallelogram and therefore throughout the plane. The Liouville theorem makes it constant, and its value at zero makes it zero. This proves the Weierstrass elliptic differential equation

Differentiating and cancelling ${\wp }^{\prime }$ wherever it is nonzero gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$; the identity extends across its isolated zeros by holomorphy. Write $\wp =z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$. For $n\ge 3$, compare coefficients of $z^{2n-2}$:The denominator is positive and rational, and $c_1=3G_4,c_2=5G_6$. Induction therefore gives a polynomial with positive rational coefficients on every monomial that occurs. Since $G_{2k}=c_{k-1}/(2k-1)$, the same holds for every $k>3$. This is the positive polynomial recurrence for lattice Eisenstein sums; for examplePositive coefficients do not assert that the complex values of the lattice sums themselves are positive.

An entire function that is doubly periodic is bounded: its absolute value has a finite maximum on the compact square $[0,1]+i[0,1]$, and periodic translations carry every point of $\mathbb{C}$ into that square. The Liouville theorem therefore proves every such analytic function is constant.

For the zero-pole count, suppose the meromorphic function is not identically zero; otherwise isolated zero multiplicities are not defined. It has only finitely many zeros and poles modulo the lattice $\Lambda =\mathbb{Z}+i\mathbb{Z}$, because a compact square can be covered by finitely many local meromorphic neighbourhoods. Choose a translated unit square with no zeros or poles on its boundary. The logarithmic derivative $f^{\prime }/f$ is doubly periodic, so integrals on opposite sides cancel, with their opposite orientations. The argument principle givesBoth counts include multiplicities. Each half-open fundamental parallelogram contains exactly one representative of each lattice orbit, and translation by a lattice vector preserves the multiplicity of a zero or pole. Consequently the counts agree in the prescribed half-open square too, even if that particular square has boundary zeros or poles. Nonzero constants have both counts zero.

For the series, let $\Lambda =\mathbb{Z}+i\mathbb{Z}$ and write it as the Weierstrass elliptic function $\wp (z)$. On a compact set avoiding $\Lambda$, choose $R$ with $|z|\le R$. For $|w|>2R$,There are $O(j)$ lattice points with $j\le |w|<j+1$, hence ${\sum }_{w\ne 0}|w{|}^{-3}$ converges by comparison with ${\sum }_j{j}^{-2}$. The Weierstrass M-test proves absolute locally uniform convergence of the tail, and the finite remaining terms define holomorphic functions off their poles. It follows that $\wp$ is a holomorphic function off $\Lambda$. Near each $w_0\in \Lambda$, separate the term $(z-w_0{)}^{-2}$; the remaining terms converge locally uniformly to a holomorphic function. Thus $\wp$ is meromorphic on $\mathbb{C}$, with a double pole at each lattice point and no other poles.

It remains to prove the required periods, rather than merely assert them from the lattice. Termwise differentiation, justified by locally uniform convergence of these holomorphic series, givesThis series is absolutely locally uniformly convergent off $\Lambda$. For either generator $\omega =1,i$, reindexing $w\mapsto w-\omega$ proves ${\wp }^{\prime }(z+\omega )={\wp }^{\prime }(z)$. Therefore $\wp (z+\omega )-\wp (z)$ has zero derivative off $\Lambda$. Its principal parts at lattice points cancel, so it extends to an entire constant $c_{\omega }$.