# Picard

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.

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