Why are the phase portrait of the simple plane pendulum and a domain coloring of sin(z) so similar? 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?
 A: The trajectories of the differential equation satisfy the implicit equation (with $v = dy/dt$) $$f(\theta,v) = \frac{v^2}{2} - \frac{g}{l} \cos \theta = A, \ A \ge -\frac{g}{l}$$
I think the white curves are the level curves of $|\sin(x+iy)|$.  These are given by 
$$g(x,y) = \cosh (2y) - \cos(2x) = B,\ B \ge 0$$
The two are related by the change of variables
$B = Al/g + 1$, $\theta = 2 x$, $v =  2 \sqrt{g/l} \sinh y$.
A: 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.
