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At an algebraic branch point, a local solution becomes meromorphic after the finite substitution $z-a=t^k$. The displayed convergent expansion has only finitely many negative terms. A square-root branch point corresponds to $k=2$; an essential singularity or logarithmic singularity need not become meromorphic under any such finite substitution.

Near an isolated essential singularity, a holomorphic function has dense image in the complex plane. If it avoided a disk about $w$, then $1/(f-w)$ would be bounded and extend across the singularity by the Riemann removable singularity theorem. If its extended value is nonzero, $f$ extends holomorphically; if zero, its finite-order zero makes $f$ a pole. Either contradicts an essential singularity.

A point where a specified complex function fails to be holomorphic, but which is approached by points of its domain. An isolated singularity has a punctured neighbourhood on which the function is holomorphic; a non-isolated singularity has other singularities arbitrarily close to it. The classification as a removable singularity, a pole, or an essential singularity applies to isolated singularities.

A meromorphic function on a punctured disk has finite pulled-back spherical area precisely when it extends to the Riemann sphere at the puncture. A removable singularity or pole gives a smooth spherical extension and finite area. At an essential singularity, the meromorphic version of the Great Picard theorem gives infinitely many preimages of all but at most two spherical targets. The area formula then makes the area infinite. The logarithmically weighted annular area divided by $\log (1/r)$ tends to the total unweighted area divided by $4\pi$, by averaging a nondecreasing area function.

Let $f$ be a holomorphic function on $0<|z-a|<R$, and suppose $a$ is an isolated essential singularity. There is a set $E\subseteq \mathbb{C}$ with at most one element such that, for every $w\notin E$ and every $0<r<R$, the equation $f(z)=w$ has infinitely many distinct solutions with $0<|z-a|<r$. Thus each nonexceptional value has preimages accumulating at $a$ in every punctured neighbourhood. The same possible exception works for all these neighbourhoods; it need not be a completely omitted value, since it could be attained only finitely often near $a$. ^[ref]

The basic example is $f(z)=e^{1/z}$ at $a=0$. It is a holomorphic function for $z\ne 0$, and its Laurent serieshas infinitely many nonzero negative-power terms, making zero an essential singularity. Its sole omitted value is zero. For the value one, its preimages are exactlyThey satisfy $e^{1/z_k}=1$ and $|z_k|=1/(2\pi |k|)\to 0$. For any $r>0$, every integer $k$ with $|k|>1/(2\pi r)$ therefore supplies a distinct solution inside $0<|z|<r$.

The diagram displays only finitely many preimages; the formula and its limit, rather than the finite picture, establish that infinitely many occur in every punctured neighbourhood. More generally, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages arewith any zero denominator excluded. Such an exclusion occurs only for $w=1$, with exactly one excluded integer $k$ for the chosen argument. The remaining preimages are distinct and tend to zero as $|k|\to \infty$.

The isolated essential singularity hypothesis is crucial. The function $f(z)=z^{-1}$ has a pole at zero; on $0<|z|<r$ its image is only $\{w:|w|>1/r\}$. It omits every value in the closed disc $|w|\le 1/r$, and each attained value has just one preimage. A pole therefore has entirely different local behaviour from an essential singularity.

The Casorati-Weierstrass theorem says that the image near an essential singularity is a dense subset of the complex plane. The Great Picard theorem replaces arbitrary approximation by exact attainment, and then repeats that attainment infinitely often. This is useful when a local restriction on the image can rule out an essential singularity and reduce the possibilities to a pole or a removable singularity. For a meromorphic function viewed as a holomorphic map into the Riemann sphere, the corresponding statement has at most two exceptional sphere values, including infinity in the count; it applies when the map has an isolated essential singularity, even if poles accumulate there. Picard's 1880 memoir proves the local result in section 15, treating the cases with and without accumulating poles separately. ^[ref]

The Great Picard theorem also explains the Little Picard theorem. For a nonconstant polynomial, every value is attained by the fundamental theorem of algebra. For an entire function which is not a polynomial, the function $g(\zeta )=f(1/\zeta )$ has an essential singularity at zero: a removable singularity would force $f$ to be constant by the Liouville theorem, and a pole would force $f$ to be a polynomial. Applying the Great Picard theorem to $g$ leaves at most one finite omitted value for $f$.

An injective holomorphic function cannot have an essential singularity: Casorati-Weierstrass theorem would give values from a small punctured neighbourhood inside the open image of a disjoint neighbourhood, contradicting injectivity. Apply this at infinity to an injective entire function. A removable singularity there would make the function constant by Liouville's theorem. A pole makes it a polynomial by the Cauchy coefficient formula; its derivative has no zero, so the fundamental theorem of algebra forces degree one.

An essential singularity is an isolated singularity which is neither removable nor a pole; equivalently its Laurent expansion has infinitely many nonzero negative-power terms.

For the final unnumbered classification, the only possible finite singularities are the zeros of the denominator, namely $z=\pm i$ and $z=k\pi$ for integers $k$. At zero,so zero is removable, with extension value one. At each nonzero $k\pi$, $\sin z$ has a simple zero, the other denominator factor is nonzero and the numerator is nonzero. These points are poles of order two, with leading coefficient $(k\pi {)}^2/((k\pi {)}^2+1)$.

At $z=\pm i$, the polynomial denominator has a simple zero. Since $\sin (\pm i)=\pm i\sinh 1\ne 0$ and the numerator is $-1$, these are simple poles. There are no essential singularities on $\mathbb{C}$ and no other finite singularities.

Away from zero, singularities occur exactly when $e^{1/z}=-1$, namelyFor $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D^{\prime }(z_k)=1/z_k\ne 0$. Thus every $z_k$ is a simple pole, with residue $z_k$.

As $|k|\to \infty$, these poles accumulate at zero. Therefore zero is a non-isolated singularity. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated essential singularity; the isolated-singularity classification does not apply. The displayed poles and their accumulation point zero exhaust the finite singularities.

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.

An essential singularity is an isolated singularity that is neither removable nor a pole. In terms of the Laurent series this means $\boxed{\text{infinitely many negative-power coefficients are nonzero}}$: no finite multiplication by a power of $z$ removes the singularity.

An essential singularity is an isolated singularity that is neither a removable singularity nor a pole. In its Laurent series, infinitely many coefficients of negative powers are nonzero. The alternatives follow because no negative terms give a removable singularity, while finitely many with a lowest nonzero power give a pole.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.

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 $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

The only pole is the simple one at $z=1$, with $\boxed{{\mathrm{Res}}_1=e}$. The point $z=0$ is an essential singularity, not another pole: multiplying $e^{1/z^2}$ by a nonvanishing analytic factor cannot remove its infinite principal part. For completeness its residue can also be read fromThe coefficient of $z^{-1}$ is $-{\sum }_{k\ge 1}1/k!=1-e$. This extra residue belongs to the essential singularity; it does not change the list of poles.

The identity theorem can be expressed as the principle of isolated zeros: a holomorphic function on a connected open set is either identically zero or has isolated zeros. Locally this follows from its Taylor series: at a zero of a function which is not locally zero, factor the first nonzero power, $f(z)=(z-z_0{)}^{m}h(z)$ with $h(z_0)\ne 0$. Connectedness then rules out a nonempty open region of zeros unless the whole function vanishes.

Set $g(z)=\overline{f(\bar{z})}$. This is holomorphic on the punctured plane: reflecting and conjugating a local power series produces another power series. The absence of an essential singularity means that the singularity at zero is removable or a pole. Thus for some integer $m\ge 0$, $z^{m}f(z)$ extends holomorphically across zero; the same is true of $z^{m}g(z)$. Their differencetherefore extends to a holomorphic function on all of $\mathbb{C}$. At every $1/n$, real-valuedness gives $H(1/n)=0$. These zeros accumulate at zero, which is now an interior point of the domain of $H$. The identity theorem forces $H\equiv 0$. For $z\ne 0$ we may divide by $z^m$, yieldingThe conjugation outside $f$ is essential. Also, applying the identity theorem directly on the punctured plane would be invalid: the accumulation point zero does not belong to that domain.

Take the circle with the usual positive, counterclockwise orientation. The only singularity is at zero, and the Laurent series givesThe coefficient of $z^{-1}$ is $1/2$, so the residue theorem givesThe essential singularity at zero still has a well-defined residue; the theorem does not require it to be a pole.

For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently zero is a non-isolated singularity, as described by reciprocal cosine with accumulating poles. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos (1/z)$ itself has an essential singularity at zero, taking its reciprocal introduces this accumulation of poles.