Solve $\sin(z) = z$ in complex numbers

Question

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!

Answer 1

Using a non trivial result by Picard: $\sin(z) - z$ has an essential singularity at $\infty$. Therefore it attains all values infinitely many times with at most one exception. Since $$\sin(z + 2\pi) - (z+2\pi) = \sin(z) - z - 2\pi$$ one can shift the image over multiples of $2\pi$. By Picard's theorem there cannot be any value (including $0$) that is only attained a finite number of times since only a single exception is allowed.

Answer 2

The function $\sin z - z$ is entire and of order $1$, so if it had finitely-many zeros then the Hadamard factorization theorem would say that

$$ \sin z - z = e^{az+b} p(z), $$

where $p(z)$ is some polynomial and $a$ and $b$ are some complex numbers.

This can't be true. We know that $\sin z - z \sim e^{-iz}$ as $z\to +i\infty$ and $\sin z - z \sim e^{iz}$ as $z \to -i\infty$ for $\operatorname{Re} z = 0$, and when taken together these properties imply the absurd conclusions that, not only is $p(z)$ constant, but $\operatorname{Im} a < 0$ and $\operatorname{Im} a > 0$.

Answer 3

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 $S$!)