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Write the period lattice as $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, with ${\omega }_1,{\omega }_2$ real-linearly independent and $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. A meromorphic function $f$ is an elliptic function for this lattice if $f(z+\lambda )=f(z)$ for every $\lambda \in \Lambda$. If it has no poles, it is an entire function bounded on the closure of a fundamental parallelogram. Periodicity makes it bounded on all of $\mathbb{C}$, so the Liouville theorem proves that an elliptic function without poles is constant.

Here are the two contour identities needed for counting its zeros and their sum. Choose a positively oriented fundamental parallelogram $P$ whose boundary avoids all zeros and poles, and put $h=f^{\prime }/f$. The argument principle givesbecause $h$ is periodic and opposite edges cancel. Here $N,P_f$ count zeros and poles with multiplicity. For the weighted integral, let $J_i$ be the integral of $h$ along the edge from a vertex $c$ to $c+{\omega }_i$. Translating the opposite edges givesThe endpoint values of $f$ agree along either edge. A continuous logarithm along the edge therefore changes by $2\pi i{n}_i$, with $n_i\in \mathbb{Z}$, and $J_i=2\pi i{n}_i$. On the other hand, the residue theorem evaluates the left side as the sum of the zero positions minus the sum of the pole positions, both with multiplicity. Thus the elliptic divisor-sum identity isUnder the stated pole hypothesis there is just one pole class, represented by zero, of order $m$. ConsequentlyChanging representatives of the zero classes changes their weighted sum by an element of $\Lambda$, so the congruence is well defined.

The Weierstrass elliptic function isOn a bounded set of $z$, the summand is $O(|\omega {|}^{-3})$ for large $|\omega |$. The lattice sum of $|\omega {|}^{-3}$ converges in real dimension two, proving Normal convergence of the Weierstrass elliptic-function series on compact sets away from $\Lambda$. Hence $\wp$ is a holomorphic function there, and at zero its principal part is $z^{-2}$, so it has a double pole. Replacing $\omega$ by $-\omega$ shows that $\wp$ is even. Differentiating the normally convergent series giveswhose absolute convergence permits reindexing by any $\lambda \in \Lambda$. Thus ${\wp }^{\prime }$ is periodic and $\wp (z+\lambda )-\wp (z)$ is constant. Substituting $-z-\lambda$ for $z$ and using evenness changes that constant to its negative, so it is zero. This proves that $\wp$ is an elliptic function, with double poles precisely at the lattice points.

The preceding zero count applied to $\wp -a$ says that it has exactly two zeros modulo $\Lambda$, counted with multiplicity. Evenness pairs any zero $z$ with $-z$. If these are distinct classes they must both be simple. If they coincide, $2z\in \Lambda$, and locally $\wp (z+t)=\wp (z-t)$: the zero has even order, which the total count forces to be two. In particular such a zero cannot be a lattice point, where $\wp$ has a pole. Every finite fibre of the Weierstrass function consists of two opposite simple points or one double nonzero half-period point.

To obtain the rational representation, first consider an even elliptic function $g$. The preceding description proves that $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ has exactly the fibres $\{z,-z\}$, so $g$ defines a function $A$ of $w=\wp (z)$. Away from the branch values a local inverse of $\wp$ makes $A$ meromorphic. At a half-period $z_0$, the local expansion is $w-w_0=c{t}^2+O(t^4)$, with $c\ne 0$ because the zero has order exactly two. The Laurent series of $g(z_0+t)$ contains only even powers, and $t^2$ is a holomorphic local coordinate as a function of $w-w_0$. Thus $A$ is meromorphic at that value too. At zero the same argument uses $1/w=t^2+O(t^6)$, proving meromorphy at infinity. A meromorphic function on the Riemann sphere is a rational function: subtract its finitely many principal parts and use compactness to make the remainder constant. This proves that even elliptic functions are rational in the Weierstrass function.

For an arbitrary elliptic function, split $f=f_{+}+f_{-}$, where $f_{\pm }(z)=(f(z)\pm f(-z))/2$. The even part is $A(\wp )$. Since ${\wp }^{\prime }$ is odd and not identically zero, $f_{-}/{\wp }^{\prime }$ is an even meromorphic elliptic function, including at the zeros of ${\wp }^{\prime }$ where the quotient may have poles. It is therefore $B(\wp )$ for a rational function $B$. The elliptic function-field decomposition isThe representation is unique: its even and odd parts determine $A$ and $B$, and $\wp$ takes every value on the Riemann sphere.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

The scaling rule for the Chordal Loewner equation shows that $A_t=\sqrt{t}{A}_1$, because $r^{-1}{U}_{r^{2}t}=a\sqrt{t}$. To identify the shape, rather than infer it from scaling alone, construct the inverse map explicitly. PutThen $r_{-}<0<r_{+}$, $r_{-}{r}_{+}=-4$, and $\alpha r_{+}+(1-\alpha )r_{-}=0$. Defineusing logarithms whose arguments lie in $(0,\pi )$ on $\mathbb{H}$. Its Laurent series is $f_t(w)=w-2t/w+O(w^{-2})$, so it has the inverse hydrodynamic normalization at infinity.

On the real interval $(r_{-}\sqrt{t},r_{+}\sqrt{t})$, the first factor has argument $\pi \alpha$ and the second has argument zero. The image therefore lies on one straight ray. Its modulus increases from zero to a maximum at $w=a\sqrt{t}$, and then decreases to zero. The complementary real intervals map to the negative and positive real axes. The argument principle, or the usual conformal slit-map construction, shows that $f_t$ maps $\mathbb{H}$ conformally onto the half-plane minus that segment; the interval traverses its two sides.

The boundary walk has winding number one around every point of the half-plane off the segment: the two traversals of the slit cancel, leaving the real boundary and a large semicircle. The argument principle therefore gives exactly one preimage of each such point.

It remains to check that this is the correct Loewner chain. For $f=f_1$,Together with $f_t(w)=\sqrt{t}f(w/\sqrt{t})$, this givesthe inverse form of the Chordal Loewner equation. Uniqueness therefore identifies its slit with $A_t$. Its endpoint is $\sqrt{t}{f}_1(a)$, where $f_1(a)\ne 0$ lies in $\mathbb{H}$. This proves that the square-root Loewner driving function generates a straight slit and thata straight ray from the boundary point $0$ to infinity.

For a closed piecewise smooth path $\gamma$ avoiding $w$, its winding number isA meromorphic function has a zero of order $m>0$ at $z_0$ whenwith $g$ holomorphic; it has a pole of order $m$ at $w_0$ when $(z-w_0{)}^{m}f(z)$ extends holomorphically and nonvanishingly there.

The argument principle states that, if $f$ has no zero or pole on $\gamma$,For a positively oriented simple contour this is the number of enclosed zeros minus poles, counted with multiplicity. It follows from the residue theorem because the residue of $f^{\prime }/f$ at a zero of order $m$ is $m$, while at a pole of order $m$ it is $-m$.

LetOn the imaginary axis,Thus $p(iy)$ always lies in the open right half-plane, has no zero there, and its continuous argument has net change zero as the imaginary-axis part of a large right-half-disc contour is traversed. On the right semicircle, $p(z)/z^4\to 1$ uniformly, so the argument change tends to that of $z^4$, namely $4\pi$. The argument principle therefore givesroots in the open right half-plane, counted with multiplicity.

For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

Cauchy's formula gives $|f^{\prime }(z_0)|\le M_R/R$ for an entire function bounded by $M$ on every radius-$R$ circle. If $f$ is globally bounded, letting $R\to \infty$ gives $f^{\prime }=0$, proving Liouville's theorem.

Set $h(w)=f(1/w)$. Its limit at zero makes the singularity removable, soConsequently $f(z)=a+c_1/z+O(z^{-2})$ and $z^2{f}^{\prime }(z)\to -c_1=:b$.

The argument principle on a large circle gives number of zeros minus poles equal to the winding number of $g$, which is zero because $g\to 1$. Thus $m=n$. The functionhas removable singularities everywhere and tends to one at infinity. Liouville gives $G=1$, hence

For a nonzero elliptic function with period lattice $\Lambda$, its zeros of a function $a_j$ and poles $b_j$, repeated with their multiplicities, have equal total number and satisfy ${\sum }_j{a}_j-{\sum }_j{b}_j\in \Lambda$. Choose a fundamental parallelogram avoiding all zeros and poles on its boundary. The argument principle applied to $f^{\prime }/f$ gives equality of the numbers. Applying the residue theorem to $z{f}^{\prime }/f$ gives their difference of sums. Pairing opposite edges, the extra factors are the two lattice generators multiplied by integrals of $f^{\prime }/f$ along an edge. These integrals are integer multiples of $2\pi i$, because the endpoints have the same nonzero function value. Thus the difference of sums is a lattice element. This links zeros of a pulled-back line to the chord-and-tangent group law.