TITLE: Why are the phase portrait of the simple plane pendulum and a domain coloring of sin(z) so similar?
QUESTION [9 upvotes]: The simple plane pendulum $$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin{\theta} = 0$$
has the very perdy phase portrait

Meanwhile, a domain coloring of $\sin(z)$ in the complex plane is

Why are these so similar?

REPLY [4 votes]: The equations of the phase curves in the phase portrait of the simple plane pendulum actually correspond to different energy conservation relations:
$$
\dot{\theta}^2 - \frac{g}{l}\cos(\theta) = C_0
$$ 
And in the colored graph of $\sin(z)$ in the complex plane the lines are the lines of constant magnitude:
$$
\|\sin(x+yi)\|^2 = C
$$
which can be transformed into another form by the steps below
$$
\begin{align}
\|\sin(x)\cosh(y) + i\cos(x)\sinh(y)\|^2 &= C \\
\sin(x)^2\cosh(y)^2 +  \cos(x)^2\sinh(y)^2 &= C \\
(\sin(x)^2 + \cos(x)^2)\frac{e^{2y}+e^{-2y}}{2} +  \sin(x)^2-\cos(x)^2  &= C \\
\frac{e^{2y}+e^{-2y}}{2} -\cos(2x) &= C 
\end{align}
$$
when $y$ is not far from $0$, $\frac{e^{2y}+e^{-2y}}{2} \approx 4y^2 = (2y)^2$,so if we replace $(x,y)$ by $(u,v)$ with $u=2x, \, v=2y$, then the equation becomes
$$
v^2 -\cos(u) = C.
$$
I think this is why the two plots look so similar. When $y$ goes far from $0$, their forms may no longer be such similar.