Q: Is this an example of an equation that generates solutions? And is everything below correct?
Part $1:$
Suppose that $x$ and $z$ are independent of each other.
$$\phi(x,z)= e^{\frac{z^2}{\ln(x)}}=x,$$
where $x,z\in\Bbb C.$
$x=e^{z},e^{-z}.$
With this information I rewrote the equation as:
$\phi\big(\ln(x),x\big)=x.$
So the initial assumption that $x$ and $z$ are independent was wrong and a contradiction was reached.
Letting $z=\ln(x)\implies$ $x=x.$
In essence the whole equation is actually $x=x$ which contains all solutions.
Part $2:$
But $z$ could potentially act as a generator, and generate numbers to be fed into the equation. If $z$ is thought of like this, then it can be re-written as $z=\ln(s)$ which can now be thought of as accepting some parameter $s,$ and generating curves of the form: $$\phi_s(x)= e^{\frac{\ln^2(s)}{\ln(x)}}=x.$$
With the equation in this form, we can generate some solutions. So, if $s=1/2$ then the solutions are: $x=1/2,2.$ If $s=3/4$ then $x=3/4,4/3.$ And if $s=ni$ for some $n\in \Bbb N,$ then $x=ni,-i/n.$
Then the equation $$\phi_s(x)= e^{\frac{\ln^2(s)}{\ln(x)}}=x$$ is really just a subset of solutions to $x=x.$ How large of a subset, really comes down to what we choose as our set $s.$
If $s\in\Bbb N$ then we can generate the fixed point solutions to $\phi_s(x)$ like $x=e^2$ and $x=e^3$ and so on and so forth. Since we have $\ln^2(s)$ the solutions come in reciprocal pairs, i.e $( e^2$ and $e^{-2}).$
Part $3:$
By just looking at the function (not an equation anymore), that is, partitioning $\phi_s(x)= e^{\frac{\ln^2(s)}{\ln(x)}}$ into leaves via the generator $s\in \Bbb R^+$ we get an analytic foliation of some manifold, right? What kind of manifold?
