Recently I have been working on an approach to numerically find the roots of the equation $ 2z-\sin(2z)=0$
As you can see below, I was able to find all the roots in a specific, bounded domain
So far so good. However, I have some observations, which makes me formulate some statements, of which I am not sure if they are correct.
1. The zeros lie on a fixed curve per quadrant.
I now that if $z$ is a zero, than so will $z^*$ (the conjugate), $-z$ and $-z^*$. This limits my root-finding to one quadrant (speed up of factor 4 hooray!). But, is it true there are no zeros outside this 'curve'?
2. The zeros reach an asymptote
Considering one curve, I get the feeling that, numbering the roots $z_n$ in increasing distance from the origin,
$ \lim_{n\to\infty} Re(z_n)-Re(z_{n-1})=\pi.$
And maybe, but less convinced
$ \lim_{n\to\infty} Im(z_n)-Im(z_{n-1})=0.$
Is there proof for either of the above statements? Did I miss any exact, closed form solution to the problem?