= Great Picard theorem
{c}
{title2=$\#\{z:0<|z-a|<r,\ f(z)=w\}=\infty$}
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 $a$. https://terrytao.wordpress.com/2016/10/11/math-246a-notes-4-singularities-of-holomorphic-functions/[Terence Tao, Math 246A Notes 4 (2016), Theorem 57 and the following discussion]{ref}
The basic example is $f(z)=e^{1/z}$ at $a=0$. It is a <holomorphic function> for $z\ne0$, and its <Laurent series>
$$
e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{-n}}{n!}
$$
has infinitely many nonzero negative-power terms, making zero an <essential singularity>. Its sole <omitted value> is zero. For the value one, its <preimages> are exactly
$$
z_k=\frac1{2\pi i k}=-\frac{i}{2\pi k},\qquad k\in\mathbb Z\setminus\{0\}.
$$
They satisfy $e^{1/z_k}=1$ and $|z_k|=1/(2\pi|k|)\to0$. For any $r>0$, every <integer> $k$ with $|k|>1/(2\pi r)$ therefore supplies a distinct solution inside $0<|z|<r$.
\Image[/essential-singularity-exponential-preimages.png]
{title=A finite sample of the preimages of 1 under exp(1/z), with a magnified view of their accumulation toward the excluded origin}
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 $w\ne0$, choose an <argument> $\theta$ of $w$. All its <preimages> are
$$
z_k=\frac1{\ln|w|+i(\theta+2\pi k)},\qquad k\in\mathbb Z,
$$
with 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 $|w|\le1/r$, 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. https://numdam.org/articles/10.24033/asens.187/[Picard, Mémoire sur les fonctions entières (1880), pp. 164-166]{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 $f$.
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