Does $\sin(x+iy) = x+iy$ have infinitely many solutions? How to prove that $\sin(x+iy) = x+iy$ has infinitely many solutions? I know how to prove that $\sin(x) = x$ has only one solution, but I do not know how to extend this to complex analysis. 
 A: Following is an elementary and mundane approach which count/bound the roots of the equation
$$\sin z = z\tag{*1}$$ 
using winding number.
For any $n \in \mathbb{N}$ and $r > 0$, let


*

*$R_n = (2n+\frac32)\pi$,

*$C_n$ be the square contour centered at origin with side $2R_n$.

*$S_1, S_2, S_3, S_4$ be the following $4$ line segments whose union is $C_n$.
$$\begin{cases}
S_1 &= [ -R_n - R_ni, +R_n - R_ni ]\\
S_2 &= [ +R_n - R_ni, +R_n + R_ni ]\\
S_3 &= [ +R_n + R_ni, -R_n + R_ni ]\\
S_4 &= [ -R_n + R_ni, -R_n - R_ni ]
\end{cases}$$

*$\Delta_{S_i}$ be the charge of $\;\arg(\sin z - z)\;$ as we move $z = x + iy$ along 
the segment $S_i$ counter-clockwisely with respect to the origin.

*$\mathcal{W}_n$ be the winding number of the image of $\sin z - z$ with respect to the origin as $z$ move along $C_n$ once. By definition, it is equal to
$$\mathcal{W}_n \stackrel{def}{=}\frac{\Delta_{S_1} + \Delta_{S_2} + \Delta_{S_3} + \Delta_{S_4}}{2\pi}$$

*$\epsilon(r) = \sqrt{8}r e^{-r} + e^{-2r}\;$ and $\;\eta(r) = \frac{r}{\cosh(r) + r}$.
One the line segment $S_1 = [-R_n-R_ni,+R_n-R_ni]$,
$$\sin z - z = \frac{1}{2i}(e^{ix+R_n} - e^{-ix-R_n}) - z
= \frac{e^{R_n+ix}}{2i}\left( 1 - 2i z e^{-(R_n+ix)} - e^{-2(R_n+ix)} \right)
$$
Notice $\displaystyle\;\left| 2i z e^{-(R_n+ix)} - e^{-2(R_n+ix)} \right| 
\le \epsilon(R_n) \le \epsilon(R_0) \approx 0.1198\;$ is always very small, we find
$$\left|\Delta_{S_1} - 2R_n\right| \le 2\sin^{-1}\epsilon(R_n)$$
By a similar argument, we have
$$\left|\Delta_{S_3} - 2R_n\right| \le 2\sin^{-1}\epsilon(R_n)$$
On the line segment $S_2 = [ +R_n - R_ni, +R_n + R_ni ]$,
$$\sin z - z = \sin R_n\cosh y + \cos R_n\sinh y\, i - (R_n + i y) = -( \cosh y + R_n ) + iy$$
Since the real part never change sign, we have
$$\left|\Delta_{S_2}\right| = 2\tan^{-1}\eta(R_n)$$
By a similar argument, we have
$$\left|\Delta_{S_4}\right| = 2\tan^{-1}\eta(R_n)$$
Above discussions implies
$$
\begin{align}
\left| \mathcal{W}_n - \frac{4R_n}{2\pi} \right| 
&\le \frac{2}{\pi}\left(\sin^{-1}\epsilon(R_n) + \tan^{-1}\eta(R_n)\right)\\
&\le \frac{2}{\pi}\left(\sin^{-1}\epsilon(R_0) + \tan^{-1}\eta(R_0)\right)\\
&\approx 0.126
\end{align}
$$
Since both $\mathcal{W}_n$ and $\frac{4R_n}{2\pi} = 4n+3$ are integers, we get $\mathcal{W}_n = 4n+3$.
Since $\sin z - z$ is entire, counting multiplicity, there are $4n+3$ roots of $(*1)$  inside $C_n$.
It is easy to see


*

*$z = 0$ is a triple root for $\sin z - z$.

*there are no other roots of $(*1)$ on the $x$ and $y$ axis.

*By symmetry, if $z = x + iy$ is a root, so does $\pm x \pm y i$.


As a result, all non-zero roots of $(*1)$ are simple. There are $4n+1$ distinct roots inside each $C_n$.
A: Here is a fairly simple and geometrically intuitive argument based on the following fixed point theorem.
Lemma: Let $R \subset \mathbb C$ be a solid rectangle  and suppose that $f:R\rightarrow\mathbb C$ is continuous. If $f(R)\supset R$, then $f$ has a fixed point in $R$.
Note that the lemma holds for any compact, simply connected set. It's stated for solid rectangles to keep it as simple as possible.  By a "solid" rectangle, we mean that the interior is included.  The lemma is a consequence of Brower's fixed point theorem, though there's probably a much easier way to prove it.
Now, given a positive integer $n$, let $x_0=2n\pi+\pi/2$, let $y_0=\text{arccosh}(x_0)+1$ and let $R$ denote the rectangle
$$R = [x_0-\pi/2,x_0+\pi/2] \times [0,y_0].$$
It's fairly simple to parametrize the boundary of $R$ using the expansion
$$\sin(x+iy) = \sin (x) \cosh (y)+i \cos (x) \sinh (y).$$
In particular, given any fixed value of $y$, $\sin(x+iy)$ traces out the right half of an ellipse with semi-major axes of length $\cosh(y)$ and $\sinh(y)$. For large $y$, the ellipse is nearly a circle, since $\cosh(y)\approx\sinh(y)$, and the radius is quite large compared to the rectangle. The left and right sides of the rectangle map into the imaginary axis.  For $n=1$, the image looks like so:

As a result, that rectangle must contain a fixed point of the sine function.  Since this is true for each $n$, there are infinitely many fixed points.

We can also get more precise information on the location of the fixed points.  Here is a plot of the fixed point in the first quadrant with real part less than 100.

The curve is $z(t)=\cosh(t)+it$ and the fixed points appear to cluster on the curve.  We can see why this should be so by again examining the expansion of the complex sine function but this time restricted to the curve:
$$\sin (\cosh (t)+i t)=\cosh (t) \sin (\cosh (t))+i \sinh (t) \cos (\cosh (t)).$$
Now suppose that $t$ is chosen so that $\sinh(t)\cos(\cosh(t)) = t$, so that the sine function preserves the imaginary part. A graph should convince you that there are infinitely many such $t$.  Furthermore, since $\sinh(t)$ is so much larger than $t$, we must have that $\cos(\cosh(t)$ is very small so that $\cosh(t)\approx n\pi + \pi/2$.  In the case that $n$ is even, we get $\sin(\cosh (t))\approx 1$ so that the real part is nearly preserved as well.  Thus, we'd expect a fixed point nearby.
A: A widely applicable and elementary approach based on Banach's fixed point theorem can be found here.
A: The great Picard theorem says that $f(z) = \sin(z)-z = c \in \Bbb C$ infinitely often for all but possibly one value of $c$. Note that $f(z+2\pi) = f(z) - 2\pi$. Suppose some $c_0$ is not hit infinitely many times; then $c_0+2\pi$ is, say $f(z_k) = c_0+2\pi$ for some infinite sequence $(z_k)$. By the above functional equation, then, $f(z_k+2\pi) = c_0$, providing an infinite sequence that maps to $c_0$. So $f^{-1}(c)$ is infinite for every $c \in \Bbb C$; in particular, this is the case for $c=0$ as desired.
Note that the same argument shows that for any holomorphic function $g$ with a sort of periodicity (i.e., there's some $c, d \in \Bbb C$ such that for all $z \in \Bbb C$, $g(z+c) = g(z)+d$), $g^{-1}(z)$ is infinite for any $z \in \Bbb C$. (The great Picard theorem demands that $g$ not be polynomial; but the above periodicity phenomenon precludes this from being possible, lest it have infinitely many roots.)
A: Let $f(z) : = \sin z - z$, which if we let $z = x+iy$, we may write as $$f(z) = \frac{i}{2} \left( \exp(y-xi) - \exp(-y+xi) \right) - (x+iy).$$ For any fixed natural number $n$, choose a positive number $t$ significantly larger than $\log n$ and a closed contour $\Gamma = \Gamma_1 \cup \Gamma_2 \cup \Gamma_3 \cup \Gamma_4$ in the anticlockwise sense, where \begin{eqnarray*}
\Gamma_1 &=& \{ z \in \mathbb{C} : 2\pi n \leq x \leq 2(n+1)\pi, y=0 \}, \\
\Gamma_2 &=& \{ z \in \mathbb{C} : x = 2(n+1) \pi, 0 \leq y \leq t \}, \\
\Gamma_3 &=& \{ z \in \mathbb{C} : 2\pi n \leq x \leq 2(n+1) \pi, y =t \}, \\
\Gamma_4 &=& \{ z \in \mathbb{C} : 2\pi n, 0 \leq y \leq t \}. 
\end{eqnarray*} Then we consider the image of $\gamma$ under $f$. Indeed, we have \begin{eqnarray*}
f(\Gamma_1) &=& \{ w = u+iv : -2(n+1) \pi \leq u \leq -2\pi n, v =0 \}, \\
f(\Gamma_2) &=& \{ w = u+iv : u= -2(n+1) \pi, 0 \leq v \leq \frac{1}{2}(e^t - e^{-t}) -t \}, \\
f(\Gamma_3) &=& \left \{ w : \frac{1}{2} e^t - \left( \frac{1}{2} e^{-t} + t + 2(n+1) \pi \right) \leq \left| w \right| \leq \frac{1}{2}e^t + \left( \frac{1}{2} e^{-t} + t +2(n+1) \pi \right) \right \}, \\
f(\Gamma_4 )&=& \{ w : u = -2\pi n, 0 \leq v \leq \frac{1}{2} (e^t - e^{-t} ) -t \}. 
\end{eqnarray*} So the winding number of $f(\Gamma)$ around $w=0$ is $1$. By the argument principle, $f(z) = \sin z - z$ has one zero inside the contour $\Gamma$. Since $n$ was arbitrary, we conclude that $\sin z = z$ has infinitely many solutions. 
