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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. ^[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 equationIts 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. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ 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. ^[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 $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.

Use the extended-value convention for a normal family, which is essential here: every sequence in the holomorphic family has a subsequence converging uniformly on compact subsets either to a finite holomorphic function or to infinity. Convergence to infinity means ${\inf }_K|f_n|\to \infty$ for every compact $K\subset D$. Equivalently one uses local uniform convergence in the chordal metric. For holomorphic functions a nonconstant meromorphic limit cannot acquire a pole: near a supposed pole, the reciprocals of sufficiently large-index functions are holomorphic and nowhere zero, and converge to a holomorphic function with a zero. Hurwitz's theorem would force that reciprocal limit to vanish identically, giving the constant infinite limit instead.

Restriction proves that global normality implies normality near each point. For the converse, choose a countable cover by connected discs $U_j$ on which the family is normal, with smaller discs covering $D$ and having compact closure inside the corresponding $U_j$. Such a cover exists from the countable rational-disc basis and local normality. Starting with any sequence, extract a subsequence converging on $U_1$, then a further subsequence converging on $U_2$, and so on. The diagonal sequence converges on each $U_j$. On an overlap the limits agree; a finite holomorphic limit and infinity cannot both occur there. The overlap graph of this cover is connected, since otherwise its components would partition the connected domain into disjoint open subsets. Therefore all local limits are of the same type and patch to one holomorphic function or to infinity throughout $D$. Every compact set has a finite cover by the smaller discs, so convergence on their compact closures gives uniform convergence on that compact set. This proves that normality is a local property.

For the counterexample take the unit disc and the family $f_n(z)=n(1+z^2/2)$. Since $|1+z^2/2|\ge 1-|z{|}^2/2>1/2$, $f_n$ tends uniformly to infinity. Any sequence from this family either has a constant subsequence or has indices tending to infinity, so the family is normal. Its derivatives are $nz$. No subsequence of these derivatives can have a finite locally uniform limit, because their values at any fixed nonzero point grow without bound; nor can they tend locally uniformly to infinity, since every derivative vanishes at zero. Thus a holomorphic normal family need not have a normal derivative family.

Now impose the point bound at $z_0$. For any sequence of derivatives choose corresponding functions from the original family. Normality supplies a locally uniformly convergent subsequence, and the point bound excludes the infinite limit. Let its finite holomorphic limit be $f$. The Cauchy integral formula gives convergence of derivatives uniformly on compact subsets. Explicitly, for compact $K\subset D$ choose $\rho >0$ so its closed $\rho$-neighborhood lies in a compact $L\subset D$; thenEvery sequence of derivatives therefore has a locally uniform subsequential limit. This pointwise anchoring of a holomorphic normal family proves the derivative family is normal under the stated bound.