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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 Roots of 2z-sin(2z)

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?

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I just confirmed that $Re(z_{100000}-z_{99999})=3.141592654$ (that is six more digits of $\pi$ than I had yesterday :)) – Bernhard Nov 24 '12 at 22:48
up vote 9 down vote accepted

This is only a partial answer, it only really answers the first question. It won't fit in a comment though, so I'll put it here (is this the right thing to do?).

Write $z = a+bi$. Note that $\sin(a+bi) = \sin a \cosh b +i\cos a \sinh b$. This means that $\sin z = z$ is equivalent to the system of equations $$a = \sin a \cosh b$$ $$b = \cos a \sinh b.$$ If we divide the first equation by $\cosh b$ and the second by $\sinh b$, and use $\sin^2 a + \cos^2 a = 1$, we get $$\left(\frac{a}{\cosh b}\right)^2 + \left(\frac{b}{\sinh b}\right)^2 = 1.$$ This allows us to write $a^2 = \cosh^2 b - b^2 \coth^2 b$. This curve looks like this: Image 1 (the image is using the equation that comes from $\sin 2z = 2z$, not $\sin z = z$, so you can compare it to your points; it seems to agree nicely). This answers your first question: the points are confined to a curve and you will not find any outside of it. However, it does not tell us anything about where the solutions lie on the curve.

You can do something similar by taking the system of equations and dividing by $\sin a$ and $\cos a$ and using the identity $\cosh^2 b - \sinh^2 b = 1$, which gives us $b^2 = a^2 \cot^2 a - \cos^2 a$. We can plot this together with the previous curve to get a much less nice-looking picture (I'm only plotting the positive values for $b$ here to get a less cluttered image): Image 2.
I'd be a little more skeptical of this plot than the other one, since this function behaves so erratically. This gives us a necessary condition for our roots; they have to lie on the intersections pictured. However, I can't really think of how to find precisely which intersections are roots to the equation.

Of course, if you can prove that the real parts are evenly spaced in the limit, then by considering the derivative of the first curve, it is easy to see that the limit of $\operatorname{Im}(z_n) - \operatorname{Im}(z_{n-1})$ is $0$.

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It is definitely OK to post it as an answer, I shouldn't have asked two questions in one. I asked them in order of importance. I will wait to accept to see if anyone comes up with something on the second question. Thanks for this insight! – Bernhard Nov 24 '12 at 18:18
Just some thoughts. In the limit of large $a$, the intersections of the red and blue curve get closer together. This is probably related to the period of $\cos^2 a$. – Bernhard Nov 24 '12 at 22:52
Are you able to numerically get any information about which intersections give zeros? Do you have any guesses? – Ben Nov 25 '12 at 0:43
Hmm, I have some difficulties plotting my dots and the two functions in a single graph, but it may be that every fourth intersection is a zero. I am not sure about this. – Bernhard Nov 25 '12 at 9:22

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