If \(f\) is meromorphic on a punctured neighborhood \(\left\{{0 \leq {\left\lvert {z-z_0} \right\rvert} < \delta}\right\}\) and \(f\) omits 3 values at \(z_0\), then \(f\) extends meromorphically over \(z_0\).

If \(f\) is nonconstant and entire, it takes on every value in \({\mathbf{C}}\) with at most one exception. Equivalently, if \(f\) omits 2 finite values, \(f\) is constant.

Wlog, suppose the three omitted points are \(1,0\infty\) and suppose \(f: {\mathbf{C}}\to X\coloneqq{\mathbf{CP}}^1\setminus\left\{{1,0\infty}\right\}\) is holomorphic. The universal cover of \(X\) is \({\mathbb{H}}\), so \(f\) lifts to a holomorphic map to \({\mathbb{H}}\). Under the Cayley map, we can obtain a holomorphic map \(f:{\mathbf{C}}\to {\mathbb{D}}\). This makes \(f\) a bounded entire function, and thus constant by Liouville.

## Exercise

Show that \(\sin(z) = z\) has infinitely many solutions in \({\mathbf{C}}\).

Let \(f(z) = \sin(z)-z\) and apply Big Picard: \(f\) takes on every value in \({\mathbf{C}}\), except possibly for some single \(z_0\), infinitely many times. So there is at most one \(z_0\) such that there are *finitely* many solutions to \(\sin(z) - z = z_0\). If \(z_0\neq 0\), we’re done, so suppose such an exception \(z_0\) does exist whose fiber has finitely many points. Since \(z_0+2\pi \neq z_0\), there must then be infinitely many solutions to \(f(z) = z_0 + 2\pi\). But then
\begin{align*}
f(z+2\pi) = \sin(z+2\pi) - z - 2\pi = \sin(z) - z - 2\pi = (z_0 + 2\pi)-2\pi = z_0
,\end{align*}
so we can produce infinitely many points in the fiber over \(w_0\), contradicting finiteness. So no exception \(z_0\) exists, and in particular, \(z_0=0\) is not an exception so \(f(z) = 0\) has infinitely many solutions.

Suppose that \(g\) is entire and satisfies the functional equation \(g(1-z) = 1-g(z)\). Show that \(g({\mathbf{C}}) = {\mathbf{C}}\).

By Picard, \(g\) omits at most one value \(a\). Note that \(a\neq 1/2\), since \(g(a) = g(1/2) = g(1-1/2) = 1-g(1/2) = 1-g(a)\), so \(2g(a) = 1\) and \(g(a) = 1/2\). Noting that \(1-a\neq a\) for any \(a\) other than \(1/2\), we have that \(w\coloneqq 1-a\in g({\mathbf{C}})\). Then \(g(w) \coloneqq g(1-a) = 1-g(a) = 1 - w = 1-(1-a) = a\), so \(a\in g({\mathbf{C}})\) and \(g\) omits no values.