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At the origin, the Taylor series $\sin z=z+O(z^3)$ and $1-\cos z=z^2/2+O(z^4)$ giveThus $f_1$ has a pole of order three and $f_2$ a pole of order four at zero. For the third function, the Laurent serieshas infinitely many nonzero negative-power terms, so zero is an essential singularity of $f_3$.

To classify the point at infinity on the Riemann sphere, use the coordinate $w=1/z$. Thenso $f_3$ has a simple pole at infinity.

For the other two functions, an important qualification is necessary. The poles of $f_1$ at nonzero integer multiples of $\pi$, and those of $f_2$ at nonzero integer multiples of $2\pi$, approach infinity. Equivalently, their images in the $w$ coordinate accumulate at zero. Hence infinity is a non-isolated singularity of $f_1$ and $f_2$. Neither function is holomorphic on any punctured neighbourhood of infinity, so the classification of isolated singularities into removable singularities, poles and essential singularities does not apply there.

Let $a=N+\frac{1}{2}$. The useful identity on the square contour isOn the vertical edges the first term equals one; on the horizontal edges $|y|=a$ and $\sinh \pi a>1$. Thus $|1/\sin \pi z|\le 1$ on every edge. Also $|z|\ge a$ and the contour has length $8a$, so the estimation lemma givesAt every nonzero integer $k$, the residue is $(-1{)}^k/(\pi k^2)$. At zero the Laurent series isso the residue there is $\pi /6$. The positive orientation and the residue theorem therefore giveLetting $N\to \infty$ yields the alternating reciprocal-square sum

The two opens are $U_1=\mathbb{C}\times {\mathbb{C}}^{\ast }$ and $U_2={\mathbb{C}}^{\ast }\times \mathbb{C}$; their intersection is $({\mathbb{C}}^{\ast }{)}^2$. There is just one degree-one term in the Čech cochain complex, and no degree-two term. The Čech coboundary sends $(g_1,g_2)$ to $g_2-g_1$, soThe denominator denotes the sum of the restricted function spaces.

To make the quotient explicit, expand a holomorphic function on the intersection in a normally convergent two-variable Laurent series,The terms with $m\ge 0$ extend to $U_1$. Of the remaining terms, those with $n\ge 0$ extend to $U_2$. Both subseries converge normally on their stated domains, by the coefficient bounds from the Cauchy integral formula. The unique remaining representative is the doubly negative partNo nonzero series of this form lies in the denominator, since a function on $U_1$ has no negative $z_1$ exponents and a function on $U_2$ has no negative $z_2$ exponents.

Writing $w_i=z_i^{-1}$, these representatives are exactly $w_1{w}_{2}F(w_1,w_2)$ with $F$ an entire holomorphic function on ${\mathbb{C}}^2$. To see that $F$ is entire, integrate for the coefficients on arbitrarily small product circles: $|a_{-i,-j}|\le M(r_1,r_2)r_1^i{r}_2^j$. Choosing $r_i{R}_i<1$ gives absolute convergence for $|w_i|\le R_i$, for every finite pair $R_i$. Conversely every such entire $F$ supplies a normally convergent representative on the intersection. Therefore the quotient consists of convergent doubly negative Laurent series, not merely finite Laurent polynomials:This gives the holomorphic first cohomology of punctured complex two-space; for instance $1/(z_1{z}_2)$ represents a nonzero class.

For the comparison with sheaf cohomology, $U_1,U_2$ and their intersection are Stein manifolds. They can be realized as closed complex submanifolds of affine complex spaces by adding equations $z_i{w}_i=1$ for their nonzero coordinates. Cartan theorem B makes their higher cohomology with coefficients in the sheaf of holomorphic functions vanish. Thus the cover is acyclic for ${\mathcal{O}}_X$, and the acyclic cover theorem givesThe individual cover members are Stein; their union has the nonzero cohomology just computed.

Let $\mathcal{P}={\mathcal{M}}_X/{\mathcal{O}}_X$, the sheaf of meromorphic principal parts. Exactness of taking stalks givesChoose a holomorphic coordinate $t$ vanishing at $x$. Every meromorphic germ has a finite negative tail in its Laurent series; the holomorphic tail vanishes in the quotient. HenceThis coordinate description is a vector-space identification; the intrinsic space is the quotient of germs, so a coordinate change need not preserve the displayed basis.

For each point inclusion $i_x:\{x\}\to X$, a principal part defines a local meromorphic germ near $x$, whose quotient class is zero off $x$. Gluing with zero on the complement gives a map from its skyscraper sheaf to $\mathcal{P}$. These maps induce an isomorphism on every stalk and therefore an isomorphism of sheaves:The direct sum of sheaves means the sheafification of the sectionwise direct-sum presheaf. Its sections can have infinitely many nonzero components globally, but their point supports form a locally finite family of subsets. This matches the fact that poles of a meromorphic function are locally finite. On a compact Riemann surface only finitely many points can occur; on a noncompact one a discrete infinite family is allowed.

For a global principal-part family $P$, choose local meromorphic lifts $m_i$ on a sufficiently small open cover. The differences $m_j-m_i$ are holomorphic functions and form a Čech cocycle. The connecting homomorphism sends $P$ to the resulting class in $H^1(X,{\mathcal{O}}_X)$. If this class vanishes, after refining the cover write $m_j-m_i=h_j-h_i$; then the functions $m_i-h_i$ glue to a global meromorphic function. Conversely any global lift makes the class zero. Equivalently, exactness of the given long exact sequence in sheaf cohomology saysThe connecting class is the obstruction to the Mittag-Leffler problem on a Riemann surface. It rules on the specified poles and their finite negative Laurent tails, with no additional poles allowed. When a solution exists, any two solutions differ by a global holomorphic function.

The mapping-out function of a compact H-hull is the unique conformal bijection from $H=\mathbb{H}\setminus K$ onto the complex upper half-plane with hydrodynamic normalization at infinity:The nonnegative coefficient $a_K$ is the half-plane capacity of $K$. The leading coefficient one and constant term zero remove the affine ambiguity of a half-plane map. In the compact H-hull convention, $K$ is bounded and relatively closed in $\mathbb{H}$, and its complement is a simply connected domain; boundary attachment points can be included in its closure.

We will use these standard properties explicitly: $g_K$ and its inverse extend holomorphically by Schwarz reflection principle across real intervals outside the hull's real attachment region; their real boundary restrictions are strictly increasing; and their normalized Laurent series hold in a full reflected neighbourhood of infinity. In particular, when $K$ lies in the unit disc, these extensions exist across $(-\infty ,-1)$ and $(1,\infty )$.

Write $r_{+}={\mathrm{Res}}_{z={\alpha }_{+}}g(z)$. For $\mathrm{Im}\lambda >-c$ and $\lambda \ne {\alpha }_{+}$, close $C_1$ upwards as before. The residue theorem includes the poles at $\lambda$ and ${\alpha }_{+}$, and the assumed decay in the upper region makes the arc contribution vanish. ThusIt is essential to include the apparently exceptional point $\lambda ={\alpha }_{+}$. The Laurent series there is $g(z)=r_{+}/(z-{\alpha }_{+})+g_0(z)$ with $g_0$ holomorphic. At that point the integrand is $r_{+}/(z-{\alpha }_{+}{)}^2+g_0(z)/(z-{\alpha }_{+})$, whose residue is $g_0({\alpha }_{+})$. HenceThe displayed difference is interpreted by its removable singularity. It therefore defines a holomorphic function at every point of the required half-plane, including ${\alpha }_{+}$.

Write $\mathbb{H}=\{z:\mathrm{Im}z>0\}$. A compact H-hull is a bounded, relatively closed set $A\subseteq \mathbb{H}$ for which $D_A=\mathbb{H}\setminus A$ is a simply connected domain. “Compact” refers to its compact closure in $\overline{\mathbb{H}}$; boundary points on $\mathbb{R}$ can be adjoined without changing the remaining domain.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:D_A\to \mathbb{H}$ with hydrodynamic normalization at infinity. By the Schwarz reflection principle outside a large disc, it has the Laurent seriesThe half-plane capacity is the coefficientThe empty compact H-hull has $g_A(z)=z$ and half-plane capacity zero. Nonnegativity follows from the Brownian representation of half-plane capacity proved next.

Put $u(z)=\mathrm{Im}(z-g_A(z))$ on $D_A$. This is a harmonic function, tends to zero at infinity, and is bounded. To justify the boundary behavior without requiring a locally connected space as its boundary, let $f_A=g_A^{-1}$. Its expansion at infinity implies that bounded $z$ cannot have $|g_A(z)|\to \infty$. If $z_n$ approaches a finite point of $\partial D_A$ and $g_A(z_n)$ had a subsequential limit inside $\mathbb{H}$, continuity of $f_A$ would force that point to lie inside $D_A$, a contradiction. Thus boundary degeneration under a mapping-out function gives $\mathrm{Im}{g}_A(z_n)\to 0$. Consequently $u$ extends continuously to the finite boundary with value $\mathrm{Im}z$.

Let $B$ be planar Brownian motion started at $z\in D_A$, and $\tau$ its Brownian exit time from $D_A$. This time is finite almost surely: it is at most the first time the imaginary coordinate, a one-dimensional Brownian motion, hits zero. By the Itô formula, $u(B_{t\wedge \tau })$ is a bounded martingale. The optional stopping theorem and dominated convergence theorem therefore giveBoundedness is important here: directly stopping the unbounded imaginary-coordinate martingale would require an unjustified uniform integrability assertion.

At $z=iy$, the Laurent series gives $u(iy)=a/y+O(y^{-2})$. HenceThe sign follows because the exit point lies on $\partial D_A\subseteq \overline{\mathbb{H}}$. This proves the Brownian representation of half-plane capacity.

Suppose $A\subseteq B$. Map out $A$ and let $C$ be the relatively closed image hull corresponding to $g_A(B\setminus A)$, with bounded complementary components filled if needed. Then $g_A$ maps $\mathbb{H}\setminus B$ onto $\mathbb{H}\setminus C$. Uniqueness of the mapping-out function of a compact H-hull gives $g_B=g_C\circ g_A$. Comparing their Laurent series yields the half-plane-capacity composition ruleThe Brownian representation of half-plane capacity makes the last term nonnegative, proving monotonicity of half-plane capacity:

For $r>0$, the conformal map $h(z)=r{g}_A(z/r)$ maps $\mathbb{H}\setminus rA$ onto $\mathbb{H}$ and has Laurent series $h(z)=z+r^{2}a/z+O(z^{-2})$. The uniqueness in hydrodynamic normalization at infinity givesAt $r=0$ the scaled closure is contained in $\mathbb{R}$, so the interior compact H-hull is empty and both sides are zero. This is the scaling part of scaling and translation of half-plane capacity.

For $x\in \mathbb{R}$, the conformal mapmaps $\mathbb{H}\setminus (A+x)$ onto $\mathbb{H}$ and has hydrodynamic normalization at infinity. Its Laurent series is $h(z)=z+a/z+O(z^{-2})$. Uniqueness of the mapping-out function of a compact H-hull therefore givesThis is the translation part of scaling and translation of half-plane capacity.

First make explicit the usual method of steepest descent hypotheses: the saddle point is simple, $G^{\prime }(z_0)=0$ and $g_2=G^{\prime \prime }(z_0)\ne 0$; a contour deformation leads to the two descent arms with one clockwise half-indentation around the pole; and endpoints or other contour portions contribute smaller terms. The sign of the exponential matters: $|e^{-i\lambda G}|=e^{\lambda \mathrm{Im}G}$, so decay is toward valleys of $\mathrm{Im}G$.

Use a local analytic function $w(z)$ as coordinate. Nondegeneracy allows an analytic square root after factoring the double zero of $G-G(z_0)$, so the coordinate can be chosen withThe sign of $c$ is selected by the orientation of the contour integral, with $w$ running from negative to positive real values. Matching cubic coefficients gives $d/c=-c{G}^{\prime \prime \prime }(z_0)/(6g_2)$. The transformed amplitude therefore has the Laurent seriesThe Cauchy principal value contribution of the pole on the two real arms is zero by oddness. Its clockwise semicircular indentation contributes $-i\pi F(z_0)$, since $\int dw/w=-i\pi$ there. The regular term contributes the Gaussian integral $B\sqrt{2\pi /\lambda }$. Consequently the simple saddle coinciding with a simple pole has the additive expansionThe remainder written here assumes the other contour contributions are negligible to that order. Otherwise the local result has to be supplemented by them. The next local correction is generically $O({\lambda }^{-1/2})$ times the displayed exponential. If $B=0$, it can vanish; odd Taylor terms integrate to zero between the symmetric arms. When $F(z_0)\ne 0$, the leading term gives the printed asymptotic equivalence. When $F(z_0)=0$, asymptotic equivalence to zero is not meaningful: the additive formula is still valid and a regular saddle-point approximation may lead instead.

Nondegeneracy is necessary and is not explicitly printed. For a counterexample to the unrestricted reading, take $F=1$, $z_0=0$ and $G(z)=-i{z}^4$. There is one critical point, but it is degenerate. Choose the contour from $+i\infty$ to $+\infty$, with a clockwise quarter-circle avoiding zero. Both rays are decaying valleys for $e^{-\lambda z^4}$. Their straight-ray contributions cancel, and the indentation gives $J=-i\pi /2$, not $-i\pi$. Thus the requested coefficient requires the simple saddle point and contour deformation assumptions above; the mere existence of one critical point is insufficient.

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.

On an annulus $r<|z-z_0|<R$, the Laurent theorem represents an analytic function uniquely asFor any positively oriented simple closed contour $C$ in the annulus winding once around $z_0$, the coefficients in the Laurent series areThe formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

Using the Taylor series of the exponential and series division,Thus the first three nonzero Laurent terms areThe singularities occur at $z=k\pi i$. The nearest nonzero ones are $\pm \pi i$, so this Laurent series is valid on $0<|z|<\pi$.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

A compact H-hull is a bounded subset $A\subset \mathbb{H}$, closed relative to the complex upper half-plane, such that $\mathbb{H}\setminus A$ is a simply connected domain. The word compact refers to its bounded closure in $\overline{\mathbb{H}}$; a nonempty hull need not be a compact subset of the open half-plane.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with $g_A(z)-z\to 0$ as $z\to \infty$. This hydrodynamic normalization at infinity yields a Laurent seriesThe coefficient is the half-plane capacity:

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.

On the slit plane $\mathbb{C}\setminus [-a,a]$, chooseso that $F(z)\sim z$ as $z\to \infty$. This branch cut gives, for $-a<x<a$, the boundary values

Integrate $F$ over the boundary of a large disk with a thin slit removed. The upper bank is traversed from $-a$ to $a$ and the lower bank from $a$ to $-a$, so together they contributeAt infinity the Laurent series isand hence the counterclockwise large-circle integral tends to $-\pi i{a}^2$. By the Cauchy integral theorem, the large-circle and slit-boundary contributions sum to zero. Thereforewhich yields