TITLE: Solve $\sin(z) = z$ in complex numbers
QUESTION [8 upvotes]: Show that $\sin(z) = z$ has infinitely many solutions in complex numbers.
Little Picard theorem should help, but using big Picard theorem is undesirable.
Thanks a lot!

REPLY [2 votes]: Let $k \in \mathbb{Z}$, $y \in \mathbb{R}$, $y>0$ and consider the solid rectangle $$S = \{z \in \mathbb{C} \mid 0 \leq \operatorname{Im}(z) \leq y \textrm{ and } (2k - \tfrac{1}{2})\pi \leq \operatorname{Re}(z) \leq (2k+\tfrac{1}{2})\pi \}.
$$
The function $\sin(z)$ is single valued on $S$ and maps onto to the upper half of a solid ellipse with foci $\pm 1$.  The lower, left and right sides of $S$ map onto the real interval and the top side onto the elliptic arc. (The following image may help to visualize this.)
$\sin(z)$" />
On the top side (where $\operatorname{Im}(z) = y$) the inequality $|\sin(z)| \geq \sinh(y)$ holds.  Choose $y>0$ such that $$\sinh(y) \geq \max_{z\in S}|z|$$ (note that $S$ itself depends on $y$ but $\sinh$ grows fast enough).  Then it follows that $$S \subset \sin(S).$$  If $\sin^{-1}\!: \sin(S) \to S$ is the inverse this can be written as $$S = \sin^{-1}(\sin(S)) \subset \sin(S).$$
By Brouwer's fixed point theorem $\sin^{-1}$ must have a fixed point in $S$, which will also be a fixed point of $\sin$. By symmetry of $\sin$ there is also a fixed point in $\overline{S}$, $-S$ and $-\overline{S}$.  Varying $k$ shows that $\sin$ has infinitely many fixed points.
There are also relevant complex analytic fixed point results (see here and here for example).  To name some consequences: there is at most one fixed point in the interior of $S$ and if there is one then iterating $\sin^{-1}$ on any starting point will converge to the fixed point.  (There is the technical difficulty that $\sin^{-1}$ does not map $\sin(S)$ into the interior of $S$.  Note for example that for $k=0$ there is no fixed point in the interior of $S$!)