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Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

The denominator vanishes at $z_k=1/(2\pi ik)$ for every nonzero integer $k$. These points approach zero, the denominator has a simple zero at each, and $\sin z_k\ne 0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are poles arbitrarily close to zero. The origin is a nonisolated singularity, so it has no residue defined by a Laurent series on a punctured disc. In particular it must not be classified as an isolated essential singularity merely because $e^{1/z}$ has one.

The denominator vanishes at $z_k=1/(2\pi ik)$ for every nonzero integer $k$. These points approach zero, the denominator has a simple zero at each, and $\sin z_k\ne 0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are poles arbitrarily close to zero. The origin is a nonisolated singularity, so it has no residue defined by a Laurent series on a punctured disc. In particular it must not be classified as an isolated essential singularity merely because $e^{1/z}$ has one.

Laurent's theorem states that a holomorphic function on $R<|z-a|<S$ has a unique expansion ${\sum }_{n\in \mathbb{Z}}{a}_n(z-a{)}^n$, converging absolutely there and uniformly on compact subannuli, withHere $\sin z=z(1-z^2/6+\cdots )$ has a simple zero at zero and no other zero in $|z|<\pi$. Hence $1/\sin z$ has only a simple pole there, giving the stated series with no powers below $z^{-1}$. Multiplying by $\sin z$ and comparing the constant, linear and quadratic coefficients yields

Use the coefficient integrals on the two circles. The annular region between them contains exactly the simple poles $\pi ,-\pi$, and the residue of $1/\sin z$ at each is $1/\cos (\pm \pi )=-1$. Apply the residue theorem to $1/(z^{n+1}\sin z)$ between the circles:For lower indices, set $c_n=0$ because the inner Laurent series has no such terms; the same contour argument then still applies. Since $f$ is odd, its even-index coefficients vanish. A complete explicit list for the requested range isThe subtraction accounts for the two poles crossed; using the inner coefficients unchanged outside their annulus would miss precisely these terms.

Choose a small positively oriented circle $|w|=\rho$ in $U$ within the convergence disk of the inverse series. Cauchy's coefficient formula givesLet $\gamma =g(\{|w|=\rho \})$. The biholomorphism property makes it a simple positively oriented closed curve about zero. Under $w=f(z)$, $g(w)=z$ and $dw=f^{\prime }(z)dz$, soThe last equality is (a). Apply (b), givingThis is the local Lagrange inversion theorem; the simple zero ensures that the expression being differentiated has a removable singularity and a nonzero value at zero.

The zero-prescription claim requires a closed discrete subset, meaning no accumulation in $\Omega$. If “discrete” means only that each listed point is isolated, it is false: $\{1/n:n\ge 1\}\subset \mathbb{C}$ is a discrete subset, but any holomorphic function vanishing there is identically zero by the identity theorem for holomorphic functions, and then has additional zeros. In the usual zero-set interpretation, discreteness includes local finiteness. We prove the claim with this necessary interpretation.

First choose an increasing compact exhaustion $K_j$ of $\Omega$ by relatively filled sets: every component of $\hat{\mathbb{C}}\setminus K_j$ meets $\hat{\mathbb{C}}\setminus \Omega$. Such an exhaustion is obtained by exhausting with finite unions of small closed squares lying in $\Omega$, then adding all complementary components relatively compact in $\Omega$. The filled sets remain compact in $\Omega$: a complementary component containing a point within distance $\mathrm{dist}(K_j,\mathbb{C}\setminus \Omega )$ of the complement could join that complement through a ball disjoint from $K_j$, and hence would not be filled. Their boundedness follows by using a disk containing the initial compact. Nesting can be ensured by enlarging each stage first. This is also the geometric description of the holomorphic convex hull proved below.

A finite prescribed set is realized by a finite product of linear factors. For an infinite closed discrete subset, enumerate the distinct points as $a_n$. Every compact contains only finitely many of them, so there are indices $j(n)\to \infty$ with $a_n\notin K_{j(n)}$. Finitely many initial factors can be handled without an approximation requirement. For the remaining factors, let $U_n$ be the complementary component containing $a_n$. It also contains a point $c_n\notin \Omega$ or infinity. Join $a_n$ to $c_n$ by an arc in that component, or to infinity by a ray-like arc. DefineEach $p_n$ is a holomorphic function on $\Omega$, has exactly one simple zero, at $a_n$, and no other zeros there.

The arc is disjoint from $K_{j(n)}$, and $p_n$ has a holomorphic logarithm $L_n$ near that compact. To justify the logarithm, any closed curve avoiding the arc has equal winding numbers about its finite endpoints, so the integral of $p_n^{\prime }/p_n$ around it is zero. For a ray to infinity the winding number about $a_n$ is zero. A primitive of $p_n^{\prime }/p_n$, adjusted by a constant on each component, supplies the logarithm.

The Runge theorem proved in Question 4 approximates $L_n$ uniformly on $K_{j(n)}$ by $h_n\in \mathcal{O}(\Omega )$: all needed finite poles can be chosen outside $\Omega$ because every complementary component meets its complement. Choose the error small enough thatEach $F_n$ still has exactly the simple zero $a_n$. On every fixed compact, all sufficiently late factors satisfy the summable bound above. Hence their infinite product converges locally uniformly. It has a nonzero tail because, once $|F_n-1|<1/2$, its logarithmic tails satisfy $|\log F_n|\le 2|F_n-1|$ and converge absolutely. Thus multiplying by the finitely many earlier factors gives a holomorphic function with precisely the prescribed zeros, and no additional zeros. Repeating a factor to any prescribed finite positive multiplicity gives the corresponding multiplicity version. This is prescribed zeros in a plane domain via Runge approximation, not an invocation of the existence theorem being asked for.

For the specified infinite product, put $a_k=q^{-2k-1}$. On any compact annulus $0<r\le |z|\le R$,The logarithmic-tail estimate therefore proves locally uniform convergence on ${\mathbb{C}}^{\ast }$ and nonvanishing wherever no individual factor vanishes. At a zero of one factor, the product is zero and the remaining product converges to a nonzero value. ConsequentlyAll these zeros are simple and distinct: their moduli distinguish the two strings and their indices. At $z=0$ the reciprocal factors are undefined, so the given expression has no convergence value there. The infinite sequence of zeros tending to zero also rules out continuation across that point as a holomorphic function.

Set $P(z)={\prod }_{k\ge 1}(1-q^{-2k}z)(1-q^{-2k}{z}^{-1})$. Absolute local convergence allows reindexing:Since $1-z=-z(1-z^{-1})$, $\sigma (qz)=-z\sigma (q^{-1}z)$ for every $z\ne 0$. No division by a vanishing factor is involved, so the identity also holds at all the zeros.

The argument principle states that if $h$ is meromorphic on a neighbourhood of a positively oriented simple closed contour and its interior, with neither zeros nor poles on the contour, thenwhere $N$ and $P$ count its interior zeros and poles with multiplicity. This integer is also the winding number of $h(\gamma )$ about zero.

Fix $z_0\in U$. Injectivity prevents $f$ from being constant on a neighbourhood of $z_0$, so its Taylor expansion has the formChoose a small closed disk $D\subset U$, centred at $z_0$, such that $h$ and $mh+(z-z_0)h^{\prime }$ are nonzero throughout $D$. Then $f-f(z_0)$ has its sole zero at $z_0$ in $D$, of multiplicity $m$, while $f^{\prime }$ has no zero in $D\setminus \{z_0\}$. On its boundary let $\delta =\min |f-f(z_0)|>0$.

Choose $0<|w|<\delta$. For every $0\le t\le 1$, the holomorphic function $f-f(z_0)-tw$ is nonzero on $\partial D$. By the argument principle, its number of zeros is an integer given by a contour integral varying continuously with $t$, hence remains $m$. At $t=1$ none of these zeros is $z_0$, so their derivatives are nonzero and they are simple zeros. Thus there are $m$ distinct points taking the value $f(z_0)+w$. Injectivity forces $m=1$, proving $f^{\prime }(z_0)\ne 0$ everywhere in $U$.

The complement of $D$ is nonempty and closed. Take points $w_j$ in that complement with $|z_0-w_j|\to \delta (z_0)$. This sequence is bounded, so a subsequence converges to a point $w_0\notin D$. Continuity of distance gives $|z_0-w_0|=\delta (z_0)>0$.

Put $t=(z_0-w_0)/(z-w_0)$. On the indicated exterior region $|t|<1$, and $f=1-t$ lies in the right half-plane. Its principal holomorphic logarithm is therefore the power seriesOn each compact subset of that region, $|t|\le q<1$, so the series has locally uniform convergence by comparison with $\sum q^k/k$. It does not converge uniformly on the whole region: choose $z=w_0+(z_0-w_0)/s$ with $0<s<1$. Then $t=s$; as $s\uparrow 1$, $\log (1-s)$ tends to minus infinity while each fixed partial sum stays bounded. In particular, the remainder after any fixed number of terms is unbounded.

The boundary-adapted holomorphic zero factor $E_0$ is holomorphic throughout $D$, since its only possible denominator singularity is at $w_0\notin D$. The exponential never vanishes, so its only zero is $z_0$, andOn $|z-w_0|>2\delta (z_0)$, cancellation of the first $K$ logarithmic terms gives a holomorphic logarithm of this zero-free restriction withChoose $K$ so that the final bound is smaller than the prescribed error. This logarithm is only asserted on the exterior region; a function with a zero cannot have a logarithm on all of $D$.

For each $z_n$, construct a boundary-adapted holomorphic zero factor $E_n$ whose logarithmic error is smaller than $2^{-n}$. If $L\subset D$ is compact, write $d_L=\mathrm{dist}(L,\mathbb{C}\setminus D)>0$. For all sufficiently large $n$, $2\delta (z_n)<d_L\le |z-w_n|$ for every $z\in L$. Hence $\sum \log E_n$ converges absolutely and uniformly on $L$ after discarding finitely many terms. Consequentlyis a holomorphic function. Its tail is the exponential of a convergent logarithmic sum and is nonzero. The finite initial product has precisely its prescribed simple zeros. Since $\delta (z_n)\to 0$, only finitely many $z_n$ lie in any compact subset of $D$, and this proves that $F$ has exactly the desired simple zeros.

For the final meromorphic function assertion, poles escaping to infinity need additional care: a locally finite pole sequence need not satisfy $\delta (z_n)\to 0$. Here is the required extension of the product argument. Split the distinct poles $a$ into two sets according asIn the first set, for each $\epsilon >0$ the poles with $\delta (a)\ge \epsilon$ are in a bounded closed subset lying at least $\epsilon$ from the complement. Local finiteness makes this a finite set. Thus the boundary distances tend to zero. In the second set, poles in $|a|\le R$ are at least $1/(1+R)$ from the complement, so there are finitely many of them. Thus their moduli tend to infinity.

Use the preceding factors for the first set. For the second set use the Weierstrass elementary factorsFor $|z|<|a|/2$ these have the same logarithmic-tail bound as above. If a pole has order $m$, choose the factor's error smaller than $2^{-n}/m$ and raise it to the power $m$. A possible pole at zero is handled by a finite polynomial factor. On every compact subset the resulting logarithmic tails are uniformly summable, so their product $q$ is holomorphic with zeros of exactly the pole orders and no other zeros. If $M$ is the original meromorphic function, the product $Mq$ extends holomorphically across every pole. Writing that extension as $p$ gives the concise conclusion

The assertion is false for $k\ge 2$, but true for $k=1$. For $k\ge 2$, take the jointly holomorphic right-hand side $A_{\alpha }(z,w)=k{z}^{k-1}+\alpha$. Its solution with zero initial value isAt $\alpha =0$, the value zero has multiplicity $k$ at the origin. For $\alpha \ne 0$, however, every zero is a simple zero: $f_{\alpha }^{\prime }(0)=\alpha$, and at a nonzero zero $z^{k-1}=-\alpha$,Thus there is no zero of multiplicity $k$ anywhere, even though the $k$ simple zeros stay close to zero for small $\alpha$. When $k=1$, part (a) gives exactly one zero counted with multiplicity in a small disc, so that zero is simple. Alternatively, the holomorphic implicit function theorem gives its holomorphic location as a function of $\alpha$.

For the unrestricted zero sequence, introduce the Weierstrass elementary factorsFor $|u|\le 1/2$, cancellation of the first $p$ terms of the logarithm power series gives a holomorphic logarithm of the factor:Define the canonical productFix any compact disc $|z|\le R$. Since $|z_n|\to \infty$, eventually $|z/z_n|\le 1/2$ uniformly on this disc. The tail logarithms are then bounded by $2(1/2{)}^{n+1}$, a summable bound independent of $z$. Their sum is holomorphic by locally uniform convergence of holomorphic functions, and its exponential is nowhere zero. Multiplication by the finite initial product therefore gives an entire function. Each factor has a simple zero exactly at its specified $z_n$, the exponential factor has none, and a tail beginning beyond any given compact disc is nonvanishing there. Thus $f$ has precisely the prescribed zeros and no others, each simple, and $f(0)=1$. This proves the required existence without an assumption on ${\sum }_n|z_n{|}^{-1}$.

For the final, unheaded product request, suppose now that ${\sum }_n|z_n{|}^{-1}<\infty$. On $|z|\le R$, all sufficiently late factors have $|z/z_n|\le 1/2$, so part (a) bounds their principal complex logarithms by $2R/|z_n|$. The logarithms converge absolutely and uniformly; infinite product convergence from logarithmic tails proves thatis entire and again has exactly the prescribed simple zeros. For its finite partial products,Passing to the limit provesIn particular $P$ is an entire function of exponential type. This argument uses a valid modulus estimate and does not require the false logarithm assertion in part (c).

Because $\cos 0=1$, continuity gives a disc $U$ about zero on which $\cos z\ne 0$. The function $-\tan z=(\log \cos z{)}^{\prime }$ has a holomorphic primitive $G$ on a sufficiently small disc, normalized by $G(0)=0$. Thenis holomorphic and satisfies $f(0)=1$ and $f(z{)}^2=\cos z$ on $U$.

No such square root exists throughout $|z|<2$. The point $z=\pi /2$ lies in that disc and is a simple zero of $\cos z$, since $-\sin (\pi /2)=-1$. Every zero of the square of a holomorphic function has even order of a zero of a holomorphic function, contradicting this simple zero.

A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.

A simple zero means $h(a)=0$ and $h^{\prime }(a)\ne 0$. The local factorization is $h(z)=(z-a)h_1(z)$, where $h_1$ is holomorphic and $h_1(a)=h^{\prime }(a)\ne 0$. ThusThe numerator is holomorphic near $a$, so its constant Taylor coefficient is the coefficient of $(z-a{)}^{-1}$ in the quotient. ThereforeIf $g(a)=0$, the possible pole cancels and the residue is zero; the formula does not require $g/h$ actually to have a pole. This is the residue formula for a quotient at a simple denominator zero.

The simple zero assumption gives the local factorizationwhere $k$ is holomorphic. In a small disc $k$ stays nonzero, so $q=g/k$ is holomorphic there. Expanding $q$ to first order,The residue is thereforeIf $g(a)=0$, the quotient has a removable singularity rather than a simple pole, and the same formula correctly gives zero. This is the residue formula for a quotient at a simple denominator zero.

For a polynomial $p(z)=c{\prod }_j(z-a_j)$ with $c\ne 0$, no polynomial root at zero, and $R>0$ different from every $|a_j|$, Jensen's formula gives the displayed expression. The polynomial roots are repeated according to their multiplicity of a zero: an exterior root contributes $\log |a_j|$, whereas an interior root contributes $\log R$.

For an explicit example, take the polynomialIts two simple zeros are $1/4$ and $-1/2$. The circle log-mean of a holomorphic function is thereforeAt $R=0.2$ the open disc contains no zeros of a function, at $R=0.4$ it contains $1/4$ only, and at $R=0.75$ it contains both. These three circles avoid the polynomial roots; the radii $1/4$ and $1/2$ are excluded from the displayed version.

The corresponding values of the circle log-mean of a holomorphic function and the circle geometric mean of a holomorphic function areFor example, at $R=0.4$ the contribution of the sole interior zero of a function is $\log (0.4/0.25)=\log (8/5)$; adding it to $\log (3/8)$ gives $\log (3/5)$. The second polynomial root still affects $f(0)$, but contributes no additional interior-zero term at this radius.

For $|q|>1$, define $S_q(z)={\prod }_{k\ge 0}(1-q^{-2k-1}z)(1-q^{-2k-1}{z}^{-1})$ on ${\mathbb{C}}^{\ast }$. The geometric decay gives locally uniform absolute tail convergence there. Its simple zeros are $q^{2k+1}$ and $q^{-2k-1}$ for $k\ge 0$, and reindexing gives $S_q(qz)=-z{S}_q(q^{-1}z)$. The variable zero is outside its domain, and the poles introduced by each reciprocal factor cannot be interpreted as a value of the product at zero.

// Destination: analysis.bigb

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.