= Little Picard theorem
{c}
{title2=$\#(\mathbb C\setminus f(\mathbb C))\le1$}
Every nonconstant <entire function> takes every complex value with at most one exception. Equivalently, its <image> is either $\mathbb C$ or $\mathbb C\setminus\{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a <constant function> has only one value. The <Little Picard theorem> guarantees existence of <preimages>, without requiring infinitely many <preimages> for each attained value. https://terrytao.wordpress.com/2016/10/11/math-246a-notes-4-singularities-of-holomorphic-functions/[Terence Tao, Math 246A Notes 4 (2016), Theorem 56]{ref}
For a nonconstant <polynomial> $p$, the <fundamental theorem of algebra> applied to $p(z)-w$ supplies a <preimage> for every $w\in\mathbb C$. The <complex exponential function> has precisely one <omitted value>, zero, as the explicit formula for its <preimages> shows. In contrast, the <complex sine function> has no <omitted value>. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the <quadratic equation>
$$
u^2-2iwu-1=0.
$$
Its roots have product $-1$, so they are nonzero. Every such root has a <preimage> under the <complex exponential function>, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the <periodicity of the complex sine>.
The force of the <Little Picard theorem> is that it applies to every nonconstant <entire function>, without an explicit formula. The <Liouville theorem> already gives a <dense image of a nonconstant entire function>, but a <dense subset> could still leave out many individual points. The <Little Picard theorem> limits these missing points to one. In particular, establishing two <omitted values> proves that an <entire function> is constant.
For a nonconstant <meromorphic function> on the <complex plane>, the corresponding bound is two <omitted values> in the <Riemann sphere> $\mathbb C\cup\{\infty\}$, where <poles> take the value infinity. https://numdam.org/articles/10.24033/asens.187/[Picard, Mémoire sur les fonctions entières (1880), section 7, pp. 152-153]{ref} For example,
$$
m(z)=\frac1{1-e^z}
$$
omits zero and one, and has <poles> at $2\pi i k$ for $k\in\mathbb Z$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an <entire function>, infinity is already an <omitted value> in the <Riemann sphere>, leaving at most one additional finite <omitted value>.
Proof outline: use the version of <Montel's theorem> for two fixed finite <omitted values>. https://terrytao.wordpress.com/2016/10/11/math-246a-notes-4-singularities-of-holomorphic-functions/[Tao, Math 246A Notes 4, Exercise 58]{ref} It makes the rescaled <holomorphic functions> $g_n(\zeta)=f(a+n\zeta)$ a <normal family> on the <unit disc>. The fixed value $g_n(0)=f(a)$ gives <pointwise anchoring of a holomorphic normal family>, hence a subsequence with a finite <holomorphic function> as its <locally uniform convergence> limit. The <Cauchy integral formula> bounds its <derivatives> at zero, whereas $g_n'(0)=nf'(a)$. Consequently $f'(a)=0$; arbitrary $a$ makes $f$ constant.
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