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With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

Write the Laurent series on the punctured disc as $f(z)={\sum }_{j=-\infty }^{\infty }{a}_j{z}^j$. A removable singularity means that $f$ extends holomorphically to the origin; exactly all negative-index coefficients vanish. A pole of order $k\ge 1$ means that $z^{k}f(z)$ extends with nonzero value at zero; exactly $a_{-k}\ne 0$ and every coefficient below $-k$ vanishes. An essential singularity is neither of these cases; exactly infinitely many negative-index coefficients are nonzero.

Here is a proof of the required Casorati-Weierstrass theorem consequence. If some disc of values around $w$ were missed in a punctured neighbourhood, then $|f(z)-w|\ge \epsilon >0$ there. The holomorphic function $q(z)=1/(f(z)-w)$ would be bounded, and hence have a removable singularity. If its extension had $q(0)\ne 0$, then $f=w+1/q$ would extend holomorphically. If $q(0)=0$, its zero has a finite positive order, since $q$ is not identically zero; thus $f$ would have a pole. Both contradict an essential singularity. Therefore every sufficiently small punctured neighbourhood has values arbitrarily close to every $w$. Choosing $0<|z_n|<\min (r/2,1/n)$ with $|f(z_n)-w|<1/n$ provesThe open mapping theorem states that a nonconstant holomorphic function on a domain maps open sets to open sets. If $f$ were injective with an essential singularity, choose a small open disc $U$ away from zero and a point $z_0\in U$. Its open image contains a disc about $w=f(z_0)$. The sequence just constructed has $z_n\to 0$ outside $U$, but eventually $f(z_n)\in f(U)$. Thus some $y_n\in U$ has $f(y_n)=f(z_n)$ although $y_n\ne z_n$, contradicting injectivity. An injective holomorphic function cannot have an isolated essential singularity.

For an injective entire function $g$, apply this to $f(z)=g(1/z)$. If its singularity at zero were removable, $g$ would be bounded outside a large disc, hence bounded everywhere and constant by Liouville's theorem. This contradicts injectivity. It must therefore be a pole, say of order $k$, so $|g(z)|\le C|z{|}^k$ for large $|z|$. For any Taylor coefficient of index $j>k$, the Cauchy coefficient formula gives $|a_j|\le C{R}^{k-j}$ on arbitrarily large circles; hence $a_j=0$. Thus $g$ is a nonconstant polynomial. Its derivative never vanishes by the allowed injectivity fact. If its degree exceeded one, the fundamental theorem of algebra would give a zero of this nonconstant derivative. Therefore injective entire functions are affine:

The printed proximity integral is missing a factor $1/(2\pi )$. Denote that literal integral by $m_{\mathrm{print}}$ and write $\overline{m}=m_{\mathrm{print}}/(2\pi )$. The correct Annular spherical Jensen formula, with the printed definitions of $N$ and $T$, isThe following derivation proves this identity and also isolates the normalization issue.

For the chordal metric put $u_a(z)=-\log k(f(z),a)$ and let $M_a(s)=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_a(s{e}^{i\theta })d\theta$. In a finite target coordinate,For $a=\infty$ use $u_{\infty }=\frac{1}{2}\log (1+|f{|}^2)$. Differentiating away from the preimages of $a$ gives half the squared spherical derivative of a meromorphic function. At an $a$-point of local degree $d$, $u_a=-d\log |z-z_0|+O(1)$; the second distributional derivatives give the Laplacian of $\log |z-z_0|$ is $2\pi {\delta }_{z_0}$. Thus, including poles by changing spherical coordinates,Because $a$ is off the image of $|z|=1$, $u_a$ is smooth near that circle and $I(a)=M_a^{\prime }(1)$ is finite and independent of $r$.

Apply Green's second identity to the harmonic weight $w(z)=\log (|z|/r)$ on $r<|z|<1$. On the outer boundary $w=\log (1/r)$ and ${\partial }_{n}w=1$; on the inner boundary $w=0$ and ${\partial }_{n}w=-1/r$. HenceThe distributional formula makes the left side $2\pi T(r)-2\pi N(r;a)$, and $M_a(r)-M_a(1)=\overline{m}(r;a)$, proving the corrected Annular spherical Jensen formula. Initially take an inner circle avoiding $a$-points. Logarithmic singularities have continuous angular means, and a point on the inner circle has zero logarithmic counting weight, so continuity extends the identity to every $0<r\le 1$.

There is no way to absorb the missing factor into a different constant $I(a)$. For the holomorphic function $f(z)=z$ and $a=\infty$, direct integration givesThe value of $(T-m_{\mathrm{print}})/\log (1/r)$ tends to $1/2$ as $r\downarrow 0$ and to $\pi$ as $r\uparrow 1$. Therefore the literal printed identity has no constant $I(\infty )$, whereas the corrected identity has $I(\infty )=1/2$.

For the singularity criterion, define the nondecreasing normalized spherical areaTonelli's theorem applied to the logarithmic weight gives $T(e^{-l})={\int }_0^{l}s(t)dt$. Averages of a nondecreasing function converge to its supremum: for fixed $t_0<l$, $(1-t_0/l)s(t_0)\le T(e^{-l})/l\le s(l)$. Thus the limit, not just the lower limit, exists in $[0,\infty ]$ and equalsIf $f$ has a removable singularity or pole at zero, it extends as a holomorphic map to the Riemann sphere; the pulled-back spherical density is smooth and bounded on the closed disk, hence its area is finite. Conversely an essential singularity gives, by the meromorphic version of the Great Picard theorem, infinitely many preimages of every spherical target with at most two exceptions in each punctured neighborhood. The area formula givesThe formula follows by applying change of variables on inverse-function neighborhoods away from the discrete critical points, then summing with multiplicity; the countable critical values have area zero. This proves the finite spherical area criterion for an isolated singularity and the requested equivalence between a finite lower limit and a removable singularity or pole.

Let $f$ have the stated zeros but not be identically zero. The origin is an isolated singularity. If it were a removable singularity, the extended holomorphic function would have zeros accumulating at an interior point. The proof in part (c) would force the extension, and then $f$ on its domain, to be identically zero.

If the origin were a pole of order $m$, then $g(z)=z^{m}f(z)$ would have a holomorphic extension with $g(0)\ne 0$. Continuity would make $g$ nonzero on a small disc, so $f$ would have no zeros in its punctured disc, again a contradiction. The classification of isolated singularities leaves only an essential singularity at the origin. This argument applies to every function in part (d), not just the particular sine example.

Use the curvature-minus-one normalization of the hyperbolic metric:The associated distance is $d(z,w)=2\mathrm{artanh}|(z-w)/(1-\overline{w}z)|$. A connected Riemann surface has its complete canonical Poincare metric on a Riemann surface precisely when its universal cover is conformally the unit disc. By the uniformization theorem, the other simply connected covering types are the complex plane and Riemann sphere. Disc automorphisms preserve $\lambda (z)|dz|$, as direct differentiation of $(z-a)/(1-\overline{a}z)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.

First prove the needed Schwarz lemma. If $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ give $|F(z)|\le |z|$ and $|F^{\prime }(0)|\le 1$.

Write ${\psi }_a(z)=(z-a)/(1-\overline{a}z)$. Conjugating $f$ by the source and target disc automorphisms, ${\psi }_{f(a)}\circ f\circ {\psi }_a^{-1}$ fixes zero. Applying the derivative bound givesIntegrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the Schwarz-Pick theorem. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.

If both $R$ and $S$ have disc universal covers, lift $g\circ {\pi }_R$ to $\tilde{g}:\mathbb{D}\to \mathbb{D}$ through ${\pi }_S$. The disc is simply connected, so the lift exists, and it is holomorphic because the covering maps are local biholomorphisms. Both covering maps are local hyperbolic isometries. The disc contraction therefore descends to $g^{\ast }d{s}_S\le d{s}_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a hyperbolic metric, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere universal cover of $R$ to the disc cover of $S$. A bounded entire function is constant by the Liouville theorem, and a holomorphic map from the compact sphere into the disc is constant by the maximum modulus principle. Thus a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps.

For the prescribed point, Schwarz-Pick theorem gives $|{\psi }_{w_0}(f(z))|\le |{\psi }_{z_0}(z)|$. Thereforeis holomorphic after filling the removable singularity at $z_0$, and $|h|\le 1$. Inverting the target automorphism provesConversely every holomorphic $h$ with this bound gives an admissible disc map: $|{\psi }_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.

Fix $z$ and put $r=|{\psi }_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le 1$, attain them all. Thus the one-point value region of a holomorphic disc map is $|{\psi }_{w_0}(w)|\le r$. Completing the square in $|w-w_0{|}^2\le r^2|1-{\overline{w}}_{0}w{|}^2$ gives its Euclidean centre and radius:This closed disc lies strictly inside the unit disc because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.

Near zero the magnitude of the integrand is of order $t^{\mathrm{Re}z-1}$, and at infinity it is of order $t^{\mathrm{Re}z-3}$. Thus the improper integral converges exactly in the stripAt a boundary of this strip the logarithmic-variable oscillation has no limiting primitive, so conditional convergence does not enlarge it.

Use a keyhole contour about the positive real axis for $w^{z-1}/(1+w{)}^2$, taking $0<\arg w<2\pi$. The circular arcs vanish in the indicated strip. The upper and lower banks contribute $(1-e^{2\pi iz})F(z)$. The only enclosed pole is the double pole at $w=-1$, whose residue is $(z-1)e^{i\pi z}$. The residue theorem therefore givesAt $z=1$ the quotient has a removable singularity, with limit one, agreeing with ${\int }_0^{\infty }(1+t{)}^{-2}dt=1$.

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 ${\alpha }_{+}$.

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

The function $f$ has a simple pole of residue $1/2$ at every $z=k\pi i$, because $(e^{2z}-1{)}^{\prime }=2$ there. At $z=0$, the term $1/(2z)$ in $g$ cancels the principal part of $f$. For $1\le k\le m$, the summandhas residue $1/2$ at each of $z=\pm k\pi i$, so it cancels the corresponding pole of $f$. Therefore all possible singularities of $F$ in $|z|<(m+1)\pi$ are removable singularities: they are $0$ and $\pm k\pi i$ for $1\le k\le m$. After filling them in, $F$ is analytic throughout that disc.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

Put $w=z-1$. ThenThusThis is actually a Taylor series, because $z=1$ is a regular point. The point $z=0$ is a simple pole. At infinity,extends analytically to $w=0$, so infinity is a removable singularity of $f$, with extended value zero.

For the real integral, first writeLetWith $z=e^{i\theta }$,The denominator has roots $1/2$ and $2$, so only $1/2$ lies inside the contour. Its residue is $1/(3i)$, and the residue theorem givesConsequently

SetThis is an elliptic function. Write $a=g_2/20$ and $b=g_3/28$ in the expansion from part (i). At a lattice point, the coefficient of $z^{-2}$ in $F$ is $g_2-20a=0$, and its constant coefficient is $g_3-28b=0$; there are no other singular terms. Every apparent pole is therefore a removable singularity, and $F$ extends to an entire elliptic function with value zero at the lattice points.

An entire elliptic function is bounded on a fundamental parallelogram and hence everywhere. By Liouville theorem it is constant, and its value at zero is zero. Thus $F\equiv 0$, proving the Weierstrass elliptic differential equation

The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

Both numerator and denominator vanish quadratically. Factor out $x^2$ before dividing the power series:The reciprocal of the denominator is $1+x^2/2-x^4/12+O(x^6)$. HenceThe expansion describes the continuous extension at the removable singularity $x=0$, where the original quotient is undefined but its limit is one. Odd powers vanish because the quotient is even.

Divide numerator and denominator by $x$ before taking their Maclaurin series:For the reciprocal denominator, the geometric series gives $1-x^2/6+7x^4/360+O(x^6)$. Multiplying gives the Taylor expansionStrictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

The denominator factors as $(z+2-i)(z+2+i)$. Neither root cancels against the numerator, and both are simple, so the finite singularities areThis is a rational function, so it has no branch points. At infinity, putting $w=1/z$ gives $w/(1+4w+5w^2)$, analytic at zero; infinity is therefore a removable singularity with value zero.

Clear the apparent negative power of $z$ to obtain the rational functionLet $\Delta =\sqrt{a^2-1}>0$. Its denominator factors as $(z-z_{-})(z-z_{+})$, withThese are distinct simple poles. Since $(a-\Delta )(a+\Delta )=1$ and $a+\Delta >1$, we have $0<a-\Delta <1$: only $z_{-}$ lies inside the unit circle. The apparent singularity at $z=0$ in the original expression is a removable singularity, with the continued value $h(0)=0$; it is not another pole.

Using the residue formula from part (a), with denominator derivative $2z-2ia$, givesThere are no other finite singularities of the continued rational function.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.

The Taylor series of the entire function $g$ givesThe coefficient of $(z-z_0{)}^{-1}$, and therefore the residue, isIf $g(z_0)=0$, the apparent pole is a removable singularity and the residue is zero.