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(I'm new here, so I hope this question hasn't come up before)

A bit of motivation for the problem:
It is well-known that the equations $\cos(x) = \sin(x)$, $\cos(\cos(x)) = \sin(\sin(x))$, and $\cos(\cos(\cos(x))) = \sin(\sin(\sin(x)))$ all have infinitely many solutions in $\mathbb{C}$ (the first and third have infinitely many solutions in $\mathbb{R}$ as well). The proofs in these cases are elementary, but break down when applying it to further (lacking a better word) iterations. Using the fourth to illustrate:

Consider the two functions $\cos(\cos(\cos(\cos(z))))$ and $\sin(\sin(\sin(\sin(z))))$, and for convenience, let
$H(z) = \cos(\cos(\cos(\cos(z)))) - \sin(\sin(\sin(\sin(z))))$.
Also, let $V = \{z \in \mathbb{C} \, | \, H(z) = 0\}$.

The original question (while not phrased in this manner) was:
Find $V$.

It is not difficult to verify that $\not\exists z\in V$ such that $\Im(z) = 0$. To prove this, locate local extrema of the function $H(x)$ (where in an abuse of notation, I use $H(x)$ to denote the restriction of $H$ to the real numbers), and one will find that all (relative) maximum and minimum values of the function $H(x)$ are strictly positive. In fact, one can prove that $H(x) \geq \frac{1}{10} > 0$, $\forall x \in \mathbb{R}$. There is a sharper estimate, but this is sufficient for our purposes, and proves that there are no real solutions to the equation $H(x) = 0$.

Having proved that there exist no real solutions, the question now becomes:
Is it possible to analytically (i.e. without numerical methods) prove that $V$ is non-empty?

My first instinct was to try to use the Argument Principle, as applied either to a ball of radius $n\in\mathbb{N}$ centered at $0$, or a rectangle of side-length $n$ centered at $0$, but I'm not sure if those integrals can be computed explicitly (even as contour integrals).

Note: It would be elementary to write an algorithm based on Newton's Method or some improvement thereof and attempt to find roots. But stability is an issue if you are far from a root.

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    Sorry...To clarify, it should have said "infinitely many solutions", instead of "infinitely many real solutions", as Colin pointed out. It is actually an isolated instance that in the first and third cases, there are infinitely many real solutions, as for any other iterations, there are no real solutions.2012-04-13
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    Since your function $H$ is entire, this is essentially just showing that its 'omitted value' isn't $0$ (or equivalently, that it's not of the form $e^{f(z)}$ for some $f$), but unfortunately my analysis background isn't strong enough to go much past Picard's theorem for this...2012-04-13
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    @Steven That was something I had thought of. $H$ is indeed analytic and entire, and has an essential singularity at the point at infinity. If we express this as $H(z) = e^{f(z)}$ for some $f$, I believe we would need to ensure that $f$ is analytic. Thanks for the suggestion. I'll update the question if I make any progress.2012-04-13
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    One can sharpen the condition (without analytical methods). Consider the vanishing of $H^2$ too to see, that its necessary for $cos(cos(cos(cos(z))))$ and $sin(sin(sin(sin(z))))$ to coincide, to have the same value $\pm i\sqrt{1/2}$, i.e. $V = \{z|cos(cos(cos(cos(z)))) = \pm i\sqrt{1/2} = sin(sin(sin(sin(z))))\}$. But I don't know how to go further.2012-04-14
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    It appears that there are infinitely many roots on the line $\operatorname{Re} z=\pi/2$. In this case the function $H(z)=\cos^4 z −\sin^4 z$ becomes $H(y)=\cos^3(\sinh y)−\sin^3(\cosh y)$, where the exponent denotes repeated application of the function.2012-04-14
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    @Antonio, if you wouldn't mind, can you expand your comment into an answer? I started to work it out, and I think you might be right, but I haven't been able to prove existence of infinitely many solutions in the case $\Re(z) = \pi/2$. Maybe I'm overlooking something simple, but I haven't been able to reach that conclusion.2012-04-15
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    @Nicholas, Unfortunately neither have I. I left the comment in the hopes someone else could complete the argument.2012-04-15
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    @Antonio, I've asked around on campus, and was to a professor who might be able to help with this question. I'll update if I find out anything useful.2012-04-20
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    Isn't the nonempty property obvious? Every complex analytical function that isn't a constant, has at least one zero. This one actually has no poles (it's a composition of two entire functions), only zeroes, so it must have a huge singularity at the infinity, but the fact remains, that it's not constant... so it must have a zero.2015-12-10
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    @orion Umm... $e^x$ is complex analytic, and non-constant, but has no zeroes. It sounds like you're referring to [Liouville's theorem](https://en.wikipedia.org/wiki/Liouville's_theorem_%28complex_analysis%29), but that requires that the function be bounded, which I doubt is the case here.2016-04-23

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