0$ for all $z\in\mathbb{C}$ then by continuity, it cannot be $\infty$ at any point in $\overline{\mathbb{C}}$ including at $\infty$ itself. Therefore Liouville's theorem can be expressed as \begin{Theorem}\label{liou_2} If $f()$ is right-unique analytic, finite at every point $z\in\overline{\mathbb{C}}$ and without a singular point at any point $z\in\mathbb{C}$ then $f()$ is constant $\in\mathbb{C}$. \end{Theorem} Now suppose that $f()$ is right-unique, analytic and none of its values are equal to $w\in\mathbb{C}$ at any point $z\in\overline{\mathbb{C}}$ and $f()$ has no singular points with $z\in\mathbb{C}$. Then $\frac{1}{f(z)-w}$ is everywhere finite because $f(z)$ cannot approach $w$ arbitrarily closely (for otherwise at the limit point it would equal $w$ and $\overline{\mathbb{C}}$ is compact so includes all its limit points) and analytic and has no singular points for $z\in\mathbb{C}$ by Theorem \ref{liou_2}, $\frac{1}{f(z)-w}=c\in\mathbb{C}$, therefore $f(z)$ is constant $\in\overline{\mathbb{C}}$. This proves that \begin{Theorem}\label{perm}Every right-unique analytic function $f()$ without any singular points where $z\in\mathbb{C}$ reaches every value $f(z)\in\overline{\mathbb{C}}$ for some $z\in\overline{\mathbb{C}}$ unless $f()$ is a constant $\in\overline{\mathbb{C}}$. \end{Theorem} Suppose a single component analytic function maps $p$ values each to the same $q$ values $\in\overline{\mathbb{C}}$. Does every single component analytic function have to be like this, with $p$ or $q$ allowed to be $\infty$? In the two set of values, each member of a set is equivalent to any other member. An analytic function $f()$ has a single component if and only if for every pair of points $P_1$ and $P_2$ in $\overline{\mathbb{C}}\times\overline{\mathbb{C}}$ in the graph of $f()$ there is a continuous and analytic curve starting at $P_1$ and ending at $P_2$ at each point being in the graph of $f()$ and not including any singular point of $f()$ i.e. every such point is connected not via singular points to every other such point within the graph of $f()$. For a multivalued single component analytic function $f()$ it is possible to have a circuit in which the $z$ value is returned to but $w$ comes back to a different value. That gives rise to an equation of type \eqref{eq9}. As the circuit is reduced in size, at some points the final value reached will suddenly change and will eventually will suddenly equal the original value. It suddenly changes where the curve crosses a singular point of which there can be many. Having found all the singular points and their associated equations relating the function values, it should be possible to, by following any combination of the paths in any order allowing repetition, each of which is associated with a single singular point, to get from say $(z_1,w_1)$ to any other point $(z_1,w)$ in the graph of $f()$. This would indicate that all the equations of type \eqref{eq9} have been found. It is possible (see for example \eqref{ex5}) that there is a pair (or perhaps more) of singular points that are associated with the same transformation \eqref{eq9} or its inverse. Similarly there can be circuits that return the $w$ to the same value but $z$ returns to a different value. This gives rise to an equation of type \eqref{eq3} and is equivalent to doing the same thing for $f^{o-1}()$. There could be a finite or a countably or uncountably infinite number of singular points. See for example \eqref{nonlin} with solution \eqref{nonlinsol} that has uncountably many singular points on the unit circle. For the case when the number of singular points is finite or countably infinite, this leads to the graph of $f()$ being described as a set of collections of points say $z_1,z_2 \ldots z_p, w_1,w_2\ldots w_q$ such that every one of the $z$'s is mapped to all of the $w$'s in every collection. Away from singular points, all the $z$'s are distinct and so are all the $w$'s. Therefore the positive integers $p,q$ are constants for the function $f()$, but either could be $\infty$. It may be useful to define the signature of an analytic function to be say $\{(p_1,q_1),(p_2,q_2),\ldots\}$ where each of the pairs corresponds to one component of the function. \begin{Theorem}\label{thm5.4}every analytic function reaches every value $f(z)\in\overline{\mathbb{C}}$ for some $z\in\overline{\mathbb{C}}$ unless $f()$ is a constant $\in\mathbb{C}$.\end{Theorem} \begin{proof} This follows from the corresponding property of algebraic functions ($P(z,w)=0$ always has a solution for $z$ given $w$ for any bivariate polynomial $P$) and the fact that analytic functions are continuous and are limits of sequences of algebraic functions which are all continuous. \end{proof} An interesting case occurs if the point that is the solution of such an equation approaches, under the limit, a singular point of the limit function. For example if the limit function is $f(z)= \exp(1/z)$ and the solutions approach $z=0$ as would happen if $w=0$. This works because $f(0)$ is 0 and $\infty$ i.e. both these values are attained by $f()$. \begin{Lemma}\label{nsip} An analytic function with no singular points and no inversion points is a linear function. \end{Lemma} \begin{proof} The absence of a singular point at $z=\infty$ for a function $f()$ implies a neighbourhood of $\infty$ (a large circle in the complex plane but a small circle in the Riemann Sphere) in $z$ maps in a left-unique manner locally to a neighbourhood in $w$ say centred on $w_0=f(\infty)$. If $w_0\ne\infty$ then $1/z\approx a(w-w_0)$ for very large $|z|$ therefore $dw/dz=-1/az^2\to 0$ as $z\to\infty$. Similarly if $w_0=\infty$ a small neighbourhood in $1/z$ about 0 maps to a small neighbourhood round $1/w$ at 0 so $1/z\approx b/w$ therefore $dw/dz=b$ at $(\infty,\infty)$ and if there are no singular points and no inversion points anywhere in $w(z)$ then $dw/dz$ is also everywhere finite and analytic there, so by Liouville's theorem (see for example \cite{CBV}) $dw/dz$ is constant so $w=a+bz$ where $a$ and $b$ are constants. \end{proof} \begin{Theorem}\label{nsp} An analytic function with no singular points is a bilinear function given by $f(z)=\frac{a+bz}{c+dz}$. \end{Theorem} \begin{proof} Let $f(z)=w$ be an analytic function with no singular points. Then apply a bilinear function $b()$ to $w$ such that $b(f(0))=0,b(f(1)=1,b(f(\infty)=\infty$. This can be done uniquely (see \cite{CBV} section 33). Then by Lemma~\ref{lemma3} $b(f())$ has no singular points and maps, $0\to 0$, $1\to 1$, and $\infty\to \infty$. Also $b(f())$ can have no inversion point because if some finite point $z_0\to\infty$ then $b(f())$ would be not left-unique there and would have a singular point there by Lemma~\ref{lemma6.5} contradicting the assumption. Therefore $b(f())$ satisfies the condition of Lemma~\ref{nsip} and must be a linear function i.e. $b(f(z))=\alpha+\beta z$ and $f(z)=b^{o-1}(\alpha+\beta z)$ which is also a bilinear function. \end{proof} \subsubsection{Characterising power functions} \begin{Lemma}\label{lemma5} If $p\in\mathbb{N}$ where $p>1$ then \begin{equation}\label{eq2.a}f(z)=f(e^{2\pi i/p}z)\end{equation} for all $z\in\overline{\mathbb{C}}$ for some analytic function $f()$ if and only if \begin{equation}\label{eq2.a_sol}f(z)=h(z^p)\end{equation} for all $z\in\overline{\mathbb{C}}$ where $h()$ is some other analytic function. \end{Lemma} Note: if the first step in computing $h(z)$ is to apply $z\to z^{1/p}$, all $p$ values must be included, giving a result which is a union of $p$ components. \begin{proof} Equation \eqref{eq2.a} implies all $p$ values $e^{2\pi ij/p}z$ for $0\le j\le p-1$ have the same value of $g$ and $z^p$ is the same for all these. Also the distinct sets $\{z,e^{2\pi i/p}z,e^{4\pi i/p}z,\ldots e^{(p-1)\pi i/p}z\}$ for all $z\in\overline{\mathbb{C}}$ have the union which is $\overline{\mathbb{C}}$ and are disjoint. Thus any solution of \eqref{eq2.a} on the Riemann Sphere $\overline{\mathbb{C}}$ is of the form \eqref{eq2.a_sol} and any function of this form satisfies \eqref{eq2.a} because $f(e^{2\pi i/p}z)=h((e^{2\pi i/p}z)^p)=h((e^{2\pi i/p})^pz^p)=h(z^p)=f(z)$. \end{proof} Again an argument motivating the concept of the {\em simplest solution} follows. An extra condition on $f()$ is obviously connected with an extra condition on $h()$ and vice versa because of the relationship \eqref{eq2.a_sol}. Therefore there is no extra condition on $f()$ is equivalent to saying that there is no extra condition on $h()$. There are no conditions on $h()$ at the moment, therefore this argument gives rise to the notion of the {\em simplest solution} of an equation such as \eqref{eq2.a} in which, because its general solution involves an arbitrary function $h()$ with no conditions placed on it giving rise to singular points, $h()$ will be assumed to have no singular points and therefore be a linear function. The points $z$ at which \eqref{eq2.a} requires a singular point are when the two function arguments coincide i.e. $z=e^{2\pi i/p}z$ giving $z=0$ and $\infty$. An extra condition on $f()$ modifying the behaviour surrounding the singular points at $z=0,\infty$ will require an extra condition on $h()$ also requiring a singular points at $z=0,\infty$. A singular point in $h()$ at any finite point $z_0\ne0$ implies $f()$ has singular points at all finite points $z_0^{1/p}\ne0$. Therefore the {\em simplest solution} of \eqref{eq2.a} is $f(z)=a+bz^p$. A related example is $f(z)=(z-z_0)^p$ where $p$ is a positive integer. Here the only finite singular point is at $(z_0,0)$. Introducing the variable $s$ by $s=z-z_0$, and $f^*()$ by $f^*(s)=f(z)=s^p$ then $f^*()$ satisfies \eqref{eq2.a}. Therefore expressing this in terms of $f$ using the chain of equalities \begin{equation}f(z)=f^*(s)=f^*(e^{2\pi i/p}s)=f^*(e^{2\pi i/p}(z-z_0))=f(e^{2\pi i/p}(z-z_0)+z_0)\end{equation} i.e. $f()$ satisfies \begin{equation}\label{eq7}f(z)=f(g_2(z))\text{ where }g_2(z)=e^{2\pi i/p}(z-z_0)+z_0.\end{equation}This relationship just involves the right-unique function $g_2()$. Suppose a multivalued function satisfies \begin{equation}\label{eq8}f(z)=e^{2\pi i/q}f(e^{2\pi i/p}z)\end{equation} where $p,q\in\mathbb{N}$ then this is equivalent to $f^*(z)=f^*(e^{2\pi i/p}z)$ where now $f^*(z)=(f(z))^q$ or equivalently (by Lemma~\ref{lemma5}) $f^*(z)=h(z^p)$ i.e. \begin{equation}\label{eq3_sol}f(z)=(h(z^p))^{1/q}\end{equation} for some function $h()$ and the {\em simplest solution} of \eqref{eq8} is \begin{equation}\label{ex2}f(z)=(az^p+b)^{1/q}.\end{equation} This function has finite singular points at $((-b/a)^{1/p},0)$ so if in addition $f(z)$ has no finite singular point other than at $(0,0)$ then $b=0$ and $f(z)=(az^p)^{1/q}$. If there are other singularities, equations like \eqref{eq2.a} and \eqref{eq2.a_sol} will not necessarily be exact but only asymptotically correct as the corresponding singular point is approached. For example in \eqref{ex2} if $z=(-b/a)^{1/p}+\epsilon$ then $f(z)$ can be expanded as a power series in $\epsilon$ in which terms higher than the first contribute so that the asymptotic behaviour near $((-b/a)^{1/p},0)$ is affected by the singular point at $(0,0)$. \begin{comment} Also consider $f(z)=z^{p/q}$ where $p$ and $q$ are integers. In general this will have to be described $q$ times to get back to the same value of $f(z)$. By combining the previous results it is obvious to try $g_2(z)=e^{2\pi i/p}z$ and $g_1(z)=e^{-2\pi i/q}z$. Then it is easy to show that $f(z)=g_1(f(g_2(z)))$ and if $z$ goes round the origin $q$ times, $f(z)$ will go round the origin $p$ times to come back to the same value. This relationship just involves the right-unique functions $g_1()$ and $g_2()$. Conversly, introducing the new variable $w=(z-z_0)^p$ and the new function $h()$ by \begin{equation}\label{eq5}h(w)=f(w^{1/p}+z_0)=f(z)\end{equation} then from the following series of equalities $f(g_2(z))=h((g_2(z)-z_0)^p)=h((e^{2\pi i/p}(z-z_0))^p)=h((z-z_0)^p)=h(w)$, the condition \eqref{eq2} after elimination of $f()$ in favour of $h()$ becomes the tautology $h(w)=h(w)$ so any function $f()$ of the form \eqref{eq5} satisfies \eqref{eq2}, and any solution $f()$ of \eqref{eq2} is related to a corresponding function $h()$ given by \eqref{eq5}, for which there is now no restriction so the only restriction on $h()$ is \eqref{eq5} itself i.e. $f(z)=h((z-z_0)^p)$ which has a singular point (1) at $(z_0,h(0))$ and (2) where $(z-z_0)^p$ is a singular point of $h()$. \end{comment} The ideas in Equations \eqref{eq7} and \eqref{eq1.b} can be combined by considering the solutions of \begin{equation}\label{eq4}f(z)=e^{2\pi i/q}f(e^{2\pi i/p}(z-z_0)+z_0).\end{equation} Introducing the new variable $s=z-z_0$ and the new function $f^*(s)=f(s+z_0)$ then \eqref{eq4} becomes \begin{equation}f^*(s)=e^{2\pi i/q}f^*(e^{2\pi i/p}s)\end{equation} whose general solution is $f^*(s)=(h(s^p))^{1/q}$ i.e. therefore the general solution of \eqref{eq4} is $f(z)=[h((z-z_0)^p)]^{1/q}$. As would be expected (and is justified later) the singular point(s) of $f()$ are given by \begin{enumerate}\item where the argument of the $q$-th root i.e. $h((z-z_0)^p)=0$ or $\infty$ \item where $(z-z_0)^p$ is a singular point of $h()$ \item where $z-z_0$ is a singular point of the $p$-th power function which is at 0 and at $\infty$ so $z=z_0,\infty$. \end{enumerate} For the case where $h()$ is the identity function, the second singular point no longer exists and the first and third of these singular points coincide at $z=z_0,\infty$ and $f(z)=(z-z_0)^{p/q}$ and the winding number ratio is $q:p$ in the earlier description. \begin{comment} If a small circle is described once anticlockwise in the $z$ plane around a point $z=z_0$ this will map to a small circle described once anticlockwise in the $w$ plane where $w=f(z)$ if $z_0$ is a not singular point and to an incomplete circuit or a circuit described multiple times otherwise. Equation \eqref{eq1} implies that every point $w$ is actually a member of set of $q$ points arranged equally spaced round a circle centered at the origin, and if the points $w$ are images of a small circle around $z=z_0$ they constitute $q$ small circles one about each of the set of points $w$. Now suppose the diameter of this circles in the $z$ plane increases, then the same will happen in the $w$ plane and if the circuits (now probably not precisely circles) in the $w$ plane get large enough to pass through 0, all the circuits will meet at that point. Making them larger still will result in the $w$ plane having some points interior to these circuits which are expected to be now joined up as a single circuit that loops round them $q$ times but is only described once if the circuit in the $z$ plane is described $q$ times. This is because the topology cannot change except when the circuits have the property of meeting which only happens when a single value of $f$ satisfies \eqref{eq1} which can only happen if $f=0$, but note that the corresponding value of $z$ is not fixed. This change in topology indicates that the circle now described in the $z$ plane includes a singularity which is therefore at $z=f^{-1}(0)$, and this singularity is of the type that maps a circle described once to a circuit described $q$ times so is of the type given by $z^{1/q}$. \end{comment} \section{General theory singular points} All the types of singular point so far found are of the types $q:p$ representing the winding number ratio where $p$ and $q$ are positive integers have have no common factors. These are all the types of singular points for algebraic functions. In the cases where $p$ and $q$ are finite, a singular point $(z_0,w_0)$ is a point about which if a path is traced from the starting point back to itself $q$ times in the $z$ plane this corresponds to a path in the $w$ plane described $p$ times back to itself. The most general form of equations such as \eqref{ex10}, \eqref{eq1}, \eqref{eq2.a}, \eqref{eq8} and \eqref{eq4} that describe the behaviour in the neighbourhood of a singular point seems to be \begin{equation}\label{2}f(z)=g_1(z,f(g_2(z)))\end{equation} in which $g_1$ has direct $z$ dependence in addition to its dependence on $f()$. The meaning of \eqref{2} where $g_2()$ is the identity function is that there is an associated singular point $(z_0,w_0)$ which is the point about which if a path in the $z$ plane is followed to its starting point and if the function value is followed continuously, the values of the function at each end of the path are related by \eqref{2}. This is the case where $q=1$. As will be shown, the singular point is also a point where the number of function values changes and $w_0$ is given by the different values of the function $w_0=f(z_0)$ being equal. This can be used to determine $(z_0,w_0)$. There is another version of this to describe the situation where $p=1$. In this case $g_1()$ is the identity function and the roles of $z$ and $w=f(z)$ are reversed. There is then a point $(z_0,w_0)$ about which if a continuous path is traced in the $w$ plane back to itself then the corresponding values of $z$ are related by $\eqref{2}$. The equality of these values determines the value $z_0$. In addition to these cases, for non-algebraic functions it is possible to have $q=\infty$. In this case the value of $w$ is never returned to its original value. Probably the simplest example is $w=f(z)=\ln(z)$ the inverse of the complex exponential function. This is equivalent to $z=\exp(w)=\exp(w).\exp(2\pi i)=\exp(w+2\pi i)$. Therefore $w+2\pi i=\ln(z)$ and equation~\eqref{2} is satisfied for $f()=\ln()$ and $g_2(z)=z+2\pi i$ and $g_1(z,f)=f$. Therefore the singular points are given by $z=z+2\pi i$ from lemma~\ref{lemma1} which implies $z=\infty$. This resolves the paradoxical situation with Theorem~\ref{thm1} and Lemma~\ref{nsip} and the fact that the exponential function has no finite singular points (they are at $z=\infty$ with $w=0$ and $\infty$). As in the examples above $g_1()$ and $g_2()$ are right-unique and the singular point of $f()$ is at $z=0$. \subsection{Definition and properties of singular points} In all these definitions, a neighbourhood of a point $(z,w)\in\overline{\mathbb{C}}\times \overline{\mathbb{C}}$ is an open set containing $(z,w)$ in the cartesian product topology. These results depend on general properties of mappings right-unique versus multi-valued, and left-unique versus many-to-one. These properties can be defined such that they are local to a particular point as follows. \begin{Definition}\label{def6.1} The function $f()$ is locally left-unique at $P=(z,f(z))$ if and only if there is a neighbourhood $N$ of $P$ such that for every pair $(z_1,f(z_1))$ and $(z_2,f(z_2))$ in $N$, $z_1\ne z_2\Rightarrow f(z_1)\ne f(z_2)$. \end{Definition} and likewise \begin{Definition}\label{def6.2} The function $f()$ is locally right-unique at $P=(z,f(z))$ if and only if there is a neighbourhood $N$ of $P$ such that for every pair $(z_1,f(z_1))$ and $(z_2,f(z_2))$ in $N$, $f(z_1)\ne f(z_2)\Rightarrow z_1\ne z_2$. \end{Definition} \begin{Definition}\label{def1}$f()$ has a singular point $P$ at $(z,f(z))$ if and only if for all neighbourhoods $N$ of $P$ there exists $(z_1,f(z_1))\in N$ and $(z_2,f(z_2))\in N$ such that either $[z_1\ne z_2\text{ and } f(z_1)= f(z_2)]\text{ or }[z_1=z_2\text{ and }f(z_1)\ne f(z_2)]$. \end{Definition} An equivalent statement of this is to require this condition only for all neighbourhoods of a specified neighbourhood of $P$ however small it is. This makes it clearer that the condition is a local property of the behaviour at $P$. \begin{Definition}\label{def2} This is the same as saying the condition that needs to be satisfied for the absence of a singular point of the function $f()$ at the point $P$, $(z,f(z))$ is that there exists a neighbourhood $N$ of $P$ such that \begin{equation}\forall (z_1,f(z_1)),(z_2,f(z_2))\in N [z_1=z_2\Leftrightarrow f(z_1)=f(z_2)]\end{equation} i.e. $f()$ is left-unique and right-unique within $N$. \end{Definition} Now it is easy to show that \begin{Lemma}\label{lemma6.5} A function $f()$ has a singular point at $P=(z,f(z))$ if and only if $f()$ is either not locally left-unique there or $f()$ is not locally right-unique there. \end{Lemma} \begin{proof} It is only necessary to choose the neighbourhood that is the intersection of the two neighbourhoods in definitions \ref{def6.1} and \ref{def6.2} and take the negation of the result. \end{proof} Next follows a pair of trivial yet confusing lemmas that show \begin{Lemma} If $f()$ is an analytic function that is a solution of \eqref{eq9} then $f()$ has singular points at every point $(z,w)$ that is a solution of $w=g_1(z,w)$ where $w=f(z)$.\end{Lemma} This is quite confusing because the word ``solution" is being used in different contexts and the same equation \eqref{eq9} is being used in two different ways, one to determine $f()$ and the other once $f()$ is fixed, to determine the set of singular points of $f()$. \begin{proof} Let $P$ be such a point then for every neighbourhood of $P$ there will be points where $w$ is arbitrarily close but not equal to $g_1(z,w)$. Thus by the second option of definition \ref{def1}, $P$ is a singular point of $f()$. \end{proof} And likewise \begin{Lemma}\label{lemma8} if $f()$ is an analytic function that is a solution of \eqref{eq3} then $f()$ has singular points at every point that satisfies $z=g_2(z)$.\end{Lemma} \begin{proof} likewise using the first option in \ref{def1}\end{proof} \begin{Definition}\label{def2}$f()$ has an inversion point at $(z,f(z))$ if and only if $f(z)=\infty$.\end{Definition} It is possible for a singular point to also be an inversion point e.g. $f(z)=z^{-2}$ at $z=0$. An example of an inversion point that is not a singular point is $f(z)=z^{-1}$ at $z=0$ because this function is everywhere right-unique and left-unique. The definition used in my earlier paper on algebraic functions \cite{jhn2013} includes inversion points with the singular points, and inversion points were not considered as a separate category. The reason for separating them out is for consistency in definition \ref{def1} that now works even if $f(z)=\infty$ where a neighbourhood of $\infty$ is as would be expected on the Riemann Sphere i.e. a region of the complex plane outside of a finite connected region defined by a single boundary. A topological argument involving moving $f(z_0)$ to $\infty$ where $z_0$ is a singular or inversion point suggests that the direction of traversal of $f(z)$ round a circuit surrounding $(z,f(z))$ ($P$) is the same as that of the corresponding circuit $z$ for any point $P$ in the graph of $f()$ except when $f(z_0)=\infty$ when it is reversed as the result of this circuit crossing $\infty$. \begin{Lemma}\label{lemma1} In definition \ref{def1} the location of the singular point(s) is determined by $z_1=z_2$ and $f(z_1)=f(z_2)$. \end{Lemma} \begin{proof} If a singular point $P$ for $f()$ is due to $f()$ not being left-unique, in all neighbourhoods of $P$, there exists $(z_1,f(z_1))$ and $(z_2,f(z_2))$ such that $z_1\ne z_2$ and $f(z_1)= f(z_2)$. Clearly if $z_1$ and $z_2$, which are related, are forced to satisfy $z_1=z_2$ then this has special significance. In fact this determines the location of the singular point or points. This is because if a region surrounding where $z_1$ and $z_2$ are forced to be equal is excluded, then surrounding such a point $P$ a sufficiently small neighbourhood $N$ exists such that because of $z_1\ne z_2$ not both of $z_1$ and $z_2$ can be in $N$ and the condition for a singular point fails. Therefore the singular points can only be at points $P$ given by $z_1=z_2$ where this is the only solution of $f(z_1)=f(z_2)$ which also holds at $P$. The existence of points where this is not true arbitrarily close to $P$ will guarantee that $P$ is a singular point. Likewise this will work if the singular point is due to $f()$ not being right-unique i.e. if $f()^{o-1}$ is not left-unique by swapping the roles of $z$ and $f(z)$. Thus singular points are where the number of function values changes. \end{proof} The following results relate singular behaviour to the operations of inversion, composition, addition and multiplication, and union. \begin{Lemma}\label{lemma2}$(z,f(z))$ is a singular point of $f()$ if and only if $(f(z),z)$ is a singular point of $f^{o-1}()$. \end{Lemma} \begin{Lemma}\label{lemma3}Composition with a function $h()$ that is analytic and has no singular point at a particular location implies that the singular/non-singular status of $f()$ is the same as that of $h(f())$ and $f(h())$ each at the corresponding point. \end{Lemma} \begin{proof} Suppose $h()$ is analytic and has no singular point at $(z_1,h(z_1))$ then there is a neighbourhood $N_1$ of $(z_1,h(z_1))$ such that for every pair $(z_2,h(z_2))$ and $(z_3,h(z_3))\in N_1$, $z_2=z_3\Leftrightarrow h(z_2)=h(z_3)$. Then $f()$ has no singular point at $(h(z_1),f(h(z_1)))$ if and only if there is a neighbourhood $N$ of $(h(z_1),f(h(z_1)))$ such that for every pair $(z_4,f(z_4))$ and $(z_5,f(z_5))$ in $N$, $z_4=z_5\Leftrightarrow f(z_4)=f(z_5)$. Let $N_2$ be the image of $N$ (with a typical point being $(x_1,x_2)$) under the mapping $k()$ defined by $x_1\to h^{o-1}(x_1)$, $x_2\to x_2$. This mapping is left-unique and right-unique because $h()$ is. Now let $N_3$ be the subset of $N_2$ such that the first component of each point is also in $N_1$. This will be non-empty because $N_1$ and $N_2$ are both neighbourhoods centred on a point with first component $z_1$. Then for any pair of points $(z_6,f(h(z_6))$ and $(z_7,f(h(z_7)))$ in $N_3$, $z_6=z_7\Leftrightarrow h(z_6)=h(z_7)\Leftrightarrow f(h(z_6))=f(h(z_7))$. The first equivalence is true because of the property of $h()$ and the second is true because of the property of $f()$. The existence of such a neighbourhood $N_3$ is precisely the statement that $f(h())$ has no singular point at $(z_1,f(h(z_1)))$. The other half of the theorem will be proved similarly or by considering the inverses of these functions. \end{proof}For a very similar reason \begin{Lemma}\label{lemma6} Adding or multiplying a function by another analytic function without a singular point will not alter the singular/non-singular status of the function at the corresponding point. \end{Lemma} \begin{Lemma}\label{lemma7} The only singular points of a union that are not included in one of the separate components is where at least two components intersect. \end{Lemma} These are known as intersection singular points. [needed?\begin{Lemma}\label{lemma4}If $f()$ is right-unique with a singular point at $(z_1,f(z_1))$ then $h(f())$ has a singular point at $(z_1,h(f(z_1)))$. \end{Lemma} \begin{proof} Because $f()$ is right-unique, the second option in definition \ref{def1} is not possible i.e. for all neighbourhoods $N$ of $(z_1,f(z_1))$ there exists $(z_2,f(z_2))$ and $(z_3,f(z_3))\in N$ such that $z_2\ne z_3\text{ and }f(z_2)=f(z_3)$. If $h()$ is any analytic function then $h(f(z_2))=h(f(z_3))$ where $h(f())$ is analytic, and if $h()$ is multivalued these sets are the same. Therefore for all neighbourhoods $N'$, defined as an image of $N$ under $h()$, centred on $(z_1,h(f(z_1)))$ there exists $(z_2,h(f(z_3)))$ and $(z_3,h(f(z_3)))\in N'$ where $z_2\ne z_3\text{ and }h(f(z_2))=h(f(z_3))$ implying $h(f())$ has a singular point at $(z_1,h(f(z_1)))$.\end{proof}] \begin{Lemma}\label{lemma4}If $f()$ has a singular point at $(z_1,f(z_1))$ then $h(f())$ has a singular point at $(z_1,h(f(z_1)))$. \end{Lemma} \begin{proof} $f()$ has a singular point at $(z_1,f(z_1))$ if and only if for all neighbourhoods $N$ of $(z_1,f(z_1))$ there exists $(z_2,f(z_2))$ and $(z_3,f(z_3))\in N$ such that $z_2\ne z_3\text{ and }f(z_2)=f(z_3)$ or $z_2=z_3\text{ and }f(z_2)\ne f(z_3)$. If $h()$ is any function, in the first case these conditions can be written as $[z_2\ne z_3\text{ and }h(f(z_2))=h(f(z_3))]$ where if $h()$ is multivalued these sets are the same. In the second case $[z_2=z_3\text{ and }h(f(z_2))\ne h(f(z_3))]$ if $h()$ is left-unique and if not, , . Therefore for all neighbourhoods $N'$, defined as an image of $N$ under $h()$, centred on $(z_1,h(f(z_1)))$ there exists $(z_2,h(f(z_3)))$ and $(z_3,h(f(z_3)))\in N'$ where $z_2\ne z_3\text{ and }h(f(z_2))=h(f(z_3))$ implying $h(f())$ has a singular point at $(z_1,h(f(z_1)))$.\end{proof} The importance of this result is that it is not possible to remove a singular point in a right-unique analytic function e.g. $z\to z^2$ by applying another function to the result. For example applying $z\to z^{1/2}$ gives the union $z\to \pm z$ that has an intersection singular point where these components coincide at $(0,0)$. \section{Examples of non-algebraic analytic functions and their singular and inversion points.} Analysis of behaviour in the neighbourhood of singular points similar to the above can be found for functions of a complex variable that are not algebraic as the following examples show. Returning to $f(z)=\ln(z)$, it satisfies $f(z)=f(z)+2\pi i$. Conversly $f(z)=f(z)+2\pi i$ implies, taking the $\exp$ of both sides, the identity $\exp(f(z))=\exp(f(z)+2\pi i)=h(z)$ say, for some analytic function $h(z)$ which is completely arbitrary because this imposes no condition on $h()$, therefore in general $f(z)=\ln(h(z))$. The singular point(s) of $f()$ are only where $h(z)=0\text{ or }\infty$ and at points $z$ that are singular points of $h()$. At minimum there are singular points of $f()$ only where $h(z)=0\text{ or }\infty$ when $h(z)=a+bz$ so that $h()$ has no singular points. This implies $z_0=-a/b\text{ or }\infty$ and the only fixed singular point is at $z_0=\infty$ with the other one having an arbitrary location, and the singular point by \ref{lemma1} has $w_0$ given by the solution of the single-value equation $w_0=w_0+2\pi i$ which is $w_0=\infty$. Therefore the singular points of $\ln()$ are at $(0,\infty)$ and $(\infty,\infty)$ and those of its inverse $\exp()$ are at $(\infty,0)\text{ and }(\infty,\infty)$. Consider $w=(\ln(z))^2$. Can a similar analysis for this be done? We have $w=(\ln(z)+2\pi i)^2$ then \eqref{2} is satisfied with $g_1(z,f)=(f^{1/2}+2\pi i)^2$ and $g_2(s)=s$. Note that $g_1()$ is now not right-unique. Another analysis of this sort comes from $(\ln(z))^2=(-\ln(z))^2=(\ln(z^{-1}))^2$ i.e. Equation~\eqref{2} with $g_1(z,f)=f$ and $g_2(z)=z^{-1}$, which shows that if in equation~\eqref{2} either of $g_1()$ or $g_2()$ is not right-unique, this analysis may not be unique. %Suppose $f(z)=(\ln(z))^{p/q}$ Consider $f(z)=z\ln(z)$, then \begin{equation}\label{ex4}f(z)=f(z)+2\pi iz.\end{equation} This can be represented in terms similar to \eqref{2} with single valued $g_1()$ and $g_2()$ but this time $g_1$ has direct $z$ dependence in addition to its dependence on $f()$ and $g_1(z,f)=2\pi iz+f$ and $g_2(z)=z$. Conversly from \eqref{ex4}, dividing by $z$ and taking the exponential gives the tautology $\exp(f(z)/z)=\exp(f(z)/z)$, therefore this function can be any analytic function say $h(z)$. Therefore $f(z)/z=\ln(h(z))$ and $f(z)=z\ln(h(z))$. The singular points of $f()$ are at any point where $h(z)=0\text{ or }\infty$ or at any point that is a singular point of $h()$. This gives at minimum, where $h(z)=a+bz$ with $b\ne 0$, singular points at $z=-a/b$ and $z=\infty$. It seems paradoxical to say that $z\ln(h(z))$ is the general solution of \eqref{ex4} because \eqref{ex4} just states that whatever the multivalued function $f(z)$ is, if it has any value $w$ at some point $z$, then at that point it also has the values $w+2\pi inz$ for all $n\in\mathbb{Z}$. In fact $z\ln(h(z))$ can be any analytic function $f()$ provided $h(z)=\exp(\frac{f(z)}{z})$ and \eqref{ex4} holds in the multivalued sense. Neverthless the use of the term ``general solution" in this and other cases does seem convenient. Suppose $f(z)=(\ln(z))^k$. Introduce the auxiliary function $g_2(z)=z^p$ then $f(g_2(z))=(\ln(z^p))^k=p^kf(z)$ so \eqref{2} holds with $g_1(z,f)=fp^{-k}$, and Lemma~\ref{lemma5} characterises $g_2()$. Alternatively, if only $f(g_2(z))=p^kf(z)$ and $g_2(z)=g_2(e^{2\pi i/p}z)$ then this is a set of defining equations for $f()$ involving two instances of \eqref{2} and linear functions only, one to characterise $g_2()$ and the other to define $f()$. \section{\label{sec9}The relationship between $g_2()$ and the type of singular points of $f()$ satisfying \eqref{eq3}} Consider the role played by $g_2()$ and its derivatives at an intersection point $z_1$ which is a solution of $g_2(z)=z$. This as will be seen controls to leading order the behaviour of $f(z)$ in the neigbourhood of the singular point at $z_1$ provided $f(z)$ satisfies \eqref{eq3} where $g_2()$ is as in \eqref{eq3}. First consider an arbitrary value of $g_2'(z_1)$. For $z\approx z_1$, $g_2(z)\approx g_2(z_1)+(z-z_1)g_2'(z_1)=z_1+(z-z_1)g_2'(z_1)$ therefore $f(z)\approx f(z_1+(z-z_1)g'_2(z))$. Put $z=z_1+\delta$ and treating this as an equality then $f(z_1+\delta)= f(z_1+\delta g_2'(z_1))$. A change of variable can now be made so as to relate this equation to $f(z)=f(z)+2\pi i$ with its known solution. Let $w=\ln(\delta)=\ln(z-z_1)$ and the new function $f^*()$ by $f^*(w)=f(z)$ then $f^*(w)=f^*(w+\ln g_2'(z_1))$. Now let $w=\alpha t$ and $f^+(t)=f^*(w)=f(z)$ then $f^+(t)=f^+\left(t+\frac{\ln g_2'(z_1)}{\alpha}\right)$. Then choose $\alpha$ so that $\ln(g_2'(z_1)/\alpha = 2\pi i$ i.e. $\alpha =\frac{\ln(g_2'(z_1)}{2\pi i}$ then $f^+(t)=h(\exp(t))$ i.e. \begin{equation}\label{as1}f(z)=f^*(w)=h(\exp(w/\alpha))=h\left((z-z_1)^\frac{2\pi i}{\ln(g_2'(z_1))}\right).\end{equation} This is the asymptotic behaviour of $f()$ for $z$ close to $z_1$ where $h()$ is an arbitrary analytic function. This works provided $g_2'(z_1)\ne 0$ which is clearly a special case. Now suppose $g_2'(z_1)=0$ but $g_2''(z_1)\ne 0$. Then $g_2(z)\approx g_2(z_1)+\frac{(z-z_1)^2}{2}g_2''(z_1)$ then $f()$ satisfies $f(z)=f\left(z_1+\frac{(z-z_1)^2}{2}g_2''(z_1)\right)$. Now put $k(\delta)=f(z_1+\delta)$ where as before $\delta= z-z_1$ then $k(\delta)=k(\delta^2g_2''(z_1)/2)$. Introduce $k^*()$ by $k(\delta)=k^*(\ln(\delta))$ then $k^*(\ln(\delta))=k^*(2\ln\delta+\ln(g_2''(z_1))-\ln(2))$. Introduce $w$ by $w=\ln\delta$ then $k^*(w)\approx k^*(2w)$ because as $\delta\to 0$, $|w|\to\infty$ so the other terms can be asymptotically ignored. Now introduce $k^+()$ by $k^+(\ln(x))=k^*(x)$ then $k^+(\ln w)=k^+(\ln w+\ln 2)$ so $k^+(u)=k^+(u+\ln2)$ where $u=\ln w$. Now let $t()$ be defined by $t(u\beta)=k^+(u)$ then $t(u\beta)=t(u\beta+\beta\ln(2))$. Choosing $\beta$ to be $\beta=\frac{2\pi i}{\ln(2)}$ then $t(x)=t(x+2\pi i)$ from which $t(x)=h(\exp(x))$. Undoing all these transformations now shows that $t(x)=t(u\beta)=k^+(u)=k^+(\ln(w))=k^*(w)=k(\delta)=f(z_1+\delta)=f(z)$ and $h(\exp(x))=h(\exp(\beta u))=h(\exp(\beta \ln(w)))=h(w^\beta)=h([\ln(z-z_1)]^\beta)$ so finally \begin{equation}\label{as2}f(z)=h\left([\ln(z-z_1)]^\frac{2\pi i}{\ln 2}\right)\end{equation} where this result will only be asyptotically correct as $z\to z_1$. Note that $g_2''(z_1)$ is not involved. From \eqref{as1} $g_2'(z_1)=1$ is obviously also a special case needing separate treatment. Then $g_2(z)\approx z+\frac{(z-z_1)^2}{2}g_2''(z_1)$ and the equation to be solved is $f(z)=f\left(z+\frac{(z-z_1)^2}{2}g_2''(z_1)\right)$. Putting $z=z_1+\delta$ and introducing $f^*(\delta)=f(z_1+\delta)$ gives \begin{equation}\label{fstar}f^*(\delta)=f^*\left(\delta+\frac{\delta^2}{2}g_2''(z_1)\right).\end{equation} Introduce the new variable $k$ by $k\left(\delta+\frac{\delta^2}{2}g_2''(z_1)\right)-k(\delta)=\Delta$ so that the iteration of \eqref{fstar} is transformed to an arithmetic progression, then for small $\delta$, $\frac{\delta^2}{2}g_2''(z_1)k'(\delta)=\Delta$ which can be integrated and inverted to give $\delta=-\frac{2\Delta}{kg_2''(z_1)}$. Then $f^*\left(\frac{-2\Delta}{kg_2''(z_1)}\right)= f^*\left(\frac{-2\Delta}{kg_2''(z_1)}+\frac{2\Delta^2}{k^2g_2''(z_1)}\right)$. Introducing $f^+(k)=f^*(\delta)$ this can be written in terms of $f^+()$ as $f^+(k)=f^+\left(\frac{\frac{-2\Delta}{g_2''(z_1)}}{\left(\frac{-2\Delta}{kg_2''(z_1)}+\frac{2\Delta^2}{k^2g_2''(z_1)}\right)}\right)$ which simplifies to $f^+(k)=f^+\left(\frac{k^2}{k-\Delta}\right)\approx f^+(k+\Delta)$. Let $g()$ be given by $g(l)=f^+(k)$ where $k=l/\alpha$ then $g(l)=g(l+\alpha\Delta)$ and choosing $\alpha\Delta=2\pi i$ then $g(l)=h(\exp(l))$ where $h()$ is arbitrary and this implies \begin{equation}\label{as3}f(z)=h\left(\exp\left(-\frac{4\pi i}{g_2''(z_1)(z-z_1)}\right)\right)\end{equation} asymptotically as $z\to z_1$. This result can be generalised as follows. Suppose $g_2'(z_1)=1$ and $g_2^{(n)}(z_1)=0$ for $2\le n\le m-1$ and $g_2^{(m)}(z_1)\ne 0$ for $m\ge2$. Then $g_2(z)=z+\frac{(z-z_1)^m}{m!}g_2^{(m)}(z_1)+O(z-z_1)^{m+1}$. In terms of $f^*()$ and $\delta$ as above, $f(z)=f(g_2(z))$ becomes $f^*(\delta)=f^*\left(\delta+\frac{\delta^mg_2^{(m)}(z_1)}{m!}+O(\delta^{m+1})\right)$. This can be iterated and if $k$ is chosen such that $k\left(\delta+\frac{\delta^mg_2^{(m)}(z_1)}{m!}\right)=k(\delta)+\Delta$ which can be approximated by $k'(\delta)\frac{\delta^mg_2^{(m)}(z_1)}{m!}=\Delta$ which integrates to $k(\delta)=\frac{-\Delta m!}{(m-1)\delta^{m-1}g_2^{(m)}(z_1)}$, then the iteration is an arithmetic progression and $f^*(\delta)=f^+(k)=f^+(k+\Delta)$. Therefore similarly to the above, \begin{equation}f(z)=h\left(\exp\left(\frac{-2\pi im!}{(m-1)g_2^{(m)}(z_1)(z-z_1)^{m-1}}\right)\right)\end{equation} asymptotically as $z\to z_1$. \section{Some interesting examples} Another example is \begin{equation}\label{eq18}f(z)=(f(z))^{1/2}\end{equation} with a singular point where $f(z)=0$, which is a special case of \eqref{1} in which $g_1()$ is not single valued. Taking natural logarithms twice gives \begin{equation}\ln\ln (f(z))= \ln(1/2)+\ln\ln(f(z))\end{equation} and so \begin{equation}\frac{2\pi i}{\ln(2)}\ln\ln(f(z))=-2\pi i+\frac{2\pi i}{\ln(2)}\ln\ln(f(z))\end{equation} so \begin{equation}\exp\left(\frac{2\pi i}{\ln(2)}\ln\ln(f(z))\right)\end{equation} is arbitrary so call it $h(z)$ then \begin{equation}\label{eq22}f(z)=\exp\left(\exp\left(\frac{\ln(2)}{2\pi i}\ln(h(z))\right)\right).\end{equation} The function $f()$ can only have a singular or inversion point where $h()$ has singular or inversion point(s) or where $h(z)=0$ or $\infty$ so $f(z)=0$ or $\infty$. This log-like singularity from \eqref{eq18} is characterised by the equations \begin{equation}\begin{array}{l}g_1(z)=-g_1(z)\\f(z)=g_1(f(z))\end{array}\end{equation} for the multivalued functions $f()$ and $g_1()$, where $g_1()$ is the {\em simplest solution}. If $f()$ is also the {\em simplest solution} then \begin{equation}f(z)=\exp\left(\exp\left(\frac{\ln(2)}{2\pi i}\ln(a+bz)\right)\right)\end{equation} where $a$ and $b$ are constants. Next consider \begin{equation}\label{nonlin}f(z)=f(z^2)/2.\end{equation} This is a special case of \eqref{1} in which the condition for a singular point is more complicated than for \eqref{eq3} for which the condition for a singular point would give \begin{equation}\label{sing}z=g_2(z)=z^2\end{equation} determining more than one such point i.e. $z=0,1$. The effect of the extra factor of 2 complicates this a bit but this is still clearly true. Because $g_2()$ is not left-unique, \eqref{sing} relates new singular points to other points already known to be singular points. In this example the singular points are dense on the unit circle because these are points for which $z^{(2^k)}=1$ for arbitrarily large $k$. It follows that $f(z)=f(z^2)/2=f(z^4)/4=\ldots f(z^{(2^k)})/2^k$ so if $z=re^{i\theta}$, $f(re^{i\theta})=f((re^{i\theta})^{2^k})/2^k$ for all $k>0$. For fixed $r$ and $\theta$ suppose $\theta + 2\pi p\approx2^k\theta$ where $p,k\in\mathbb{N}$ then $f(r^{(2^k)}e^{i\theta})\approx 2^kf(re^{i\theta})$. Putting $R=r^{(2^k)}$ gives $f(Re^{i\theta})\approx\frac{\ln(R)}{\ln(r)}f(re^{i\theta})$. The log dependence on $R$ behaviour at large $R$ and the positions ($z$) of the singular points may suggest the following formula \begin{equation}\label{nonlinsol}f(z)=\int_0^{2\pi}d\theta\log_2|z-e^{i\theta}|.\end{equation} for a solution of \eqref{nonlin}. Its proof is as follows \begin{equation}\begin{array}{l} f(z^2)=\int_0^{2\pi}d\theta\log_2|z^2-e^{i\theta}|=\int_0^{2\pi}d\theta\log_2\left(|z+e^{i\theta/2}||z-e^{i\theta/2}|\right)\\ =\int_0^{2\pi}d\theta\log_2|z+e^{i\theta/2}|+\int_0^{2\pi}d\theta\log_2|z-e^{i\theta/2}|\\ =2\int_0^{\pi}d\theta\log_2|z+e^{i\theta}|+2\int_0^{\pi}d\theta\log_2|z-e^{i\theta}|\\ =2\int_{\pi}^{2\pi}d\theta\log_2|z+e^{i(\theta-\pi)}|+"\\ =2\int_{\pi}^{2\pi}d\theta\log_2|z-e^{i\theta}|+"\\ =2\int_0^{2\pi}d\theta\log_2|z-e^{i\theta}|=2f(z)\end{array}.\end{equation} This example has really peculiar properties because $f(z)$ is $\infty$ on the unit circle and this appears to isolate the function into two regions that can behave somewhat independently because \eqref{sing} is satisfied for $f()$ replaced by $af()$ where $a\in\mathbb{C}$ and clearly any two different values of $a$ can be chosen inside and outside the unit circle and the solutions can be described as having a natural boundary on the unit circle. [This doesn't work for finite prescribed values because if finite values are prescribed on any closed contour the Cauchy integral formula determines a function that is everywhere analytic and finite, uniquely inside it, but does it work for the outside region?] This is an example that divides $\overline{\mathbb{C}}$ into two domains of holomorphy \cite{eom} that overlap only on the unit circle. Next follows an intriguing example where the condition for a singular point (an equation of the type \eqref{eq3}) determines two of them and the solutions found satisfy an additional equation of the type \eqref{1}. Suppose $g_2(z)=\frac{a+bz}{c+z}$. Then $g_2(z)=z$ is a quadratic equation with solutions say $z_1$ and $z_2$ such that $z_1+z_2=b-c$ and $z_1z_2=-a$ and $g_2(z)$ can be written as $g_2(z)=\frac{-z_1z_2+bz}{b-z_1-z_2+z}$. However in this case, $g_2()$ is left-unique and single valued so only two singular points arise as a result of \eqref{eq3} which becomes in this case \begin{equation}\label{ex5}f(z)=f\left(\frac{bz-z_1z_2}{b-z_1-z_2+z}\right).\end{equation} Therefore by lemma \ref{lemma8} solutions of \eqref{ex5} have singular points at $z_1$ and $z_2$. Using methods similar to those used in deriving \eqref{as1} it possible to formally derive \begin{equation}f(z)=h_k\left(\sum_{n\in\mathbb{Z}}c_n \exp\left(\frac{2\pi i(\ln(z-z_k)+2n_1\pi i)}{\ln\left(\frac{z_1-b}{z_2-b}\right)+2n\pi i}\right)\right)\end{equation} for $k=1,2$ where $h_1()$ and $h_2()$ are arbitrary functions. By trial and error, the following are possible solutions of \eqref{ex5}: \begin{equation}\label{guess}f(z)=\left(c_n\frac{z-z_1}{z-z_2}\right)^s\end{equation} where $s=\frac{2\pi i}{\ln\left(\frac{z_1-b}{z_2-b}\right)+2n\pi i}$ and $n\in \mathbb{Z}$. It is easy to show that \begin{equation}\label{res1}\frac{g_2(z)-z_1}{g_2(z)-z_2}=\frac{(z_1-b)(z_1-z)}{(z_2-b)(z_2-z)}.\end{equation} Therefore \begin{equation}f(g_2(z))=\left(c_n\frac{z-z_1}{z-z_2}\right)^s\left(\frac{b-z_1}{b-z_2}\right)^s.\end{equation} The extra factor is $\left(\frac{b-z_1}{b-z_2}\right)^s$ can be written (including all its possible values) as \begin{equation}\label{exfac}\exp(s\ln(t))=\exp\left(\frac{\ln(t)\times 2\pi i}{\ln(t)+2n\pi i}\right)=\exp\left(\left(\frac{\ln(t)+2n_1\pi i}{\ln(t)+2n\pi i}\right)\times2\pi i\right)=E_{n_1,n}\end{equation} where $n_1,n\in\mathbb{Z}$ for some specific value of $\ln(t)$ and where $t=\frac{b-z_1}{b-z_2}$. Increasing $n_1$ by 1 adds $\frac{2\pi i\times 2\pi i}{\ln(t)+2n\pi i}$ to the argument of $\exp()$ multipling the whole expression by $\exp\left(\frac{-4\pi^2}{\ln(t)+2n\pi i}\right)$ and $E_{n,n}=1$. From these it follows that $E_{n_1,n}=\exp\left(\frac{4\pi^2(n-n_1)}{\ln(t)+2n\pi i}\right)$. Therefore \begin{equation}\label{eq37} f(g_2(z))=\left(c_n\frac{z-z_1}{z-z_2}\right) ^{\frac{2\pi i}{\ln\left(\frac{z_1-b}{z_2-b}\right)+2n\pi i}} \exp\left(\frac{4\pi^2(n-n_1)}{\ln\left(\frac{b-z_1}{b-z_2}\right)+2n\pi i}\right).\end{equation} From \eqref{eq3} \begin{equation}f(z)=\exp\left(\frac{2\pi i}{\ln(t)+2n\pi i}\times\ln\left(c_n\frac{z-z_1}{z-z_2}\right)\right).\end{equation} Taking this continuously round a small circuit $C_1$ anticlockwise round $z_1$ given by $z=z_1+\epsilon e^{i\theta}$ for $0\le\theta\le 2\pi$ where $\epsilon$ is a very small positive real number gives $f(z)=\exp\left(\frac{2\pi i}{\ln(t)+2n\pi i}\ln\left(\frac{c_ne^{i\theta}}{z-z_2}\right)\right)= \exp\left(\left(\ln(c_n)+i\theta-\ln(z-z_2)\right)\frac{2\pi i}{\ln(t)+2n\pi i}\right)$ The difference over the path $C_1$ of the argument of $\exp()$ is $\frac{2\pi i.2\pi i}{\ln(t)+2n\pi i}$ so the factor associated with doing $C_1$ is $\exp\left(\frac{-4\pi^2}{\ln(t)+2n\pi i}\right)$ i.e. $f(z)$ satisfies $f(z)=f(z)\exp(\frac{-4\pi^2}{\ln(t)+2n\pi i})$. This can be applied to write \eqref{eq37} as \eqref{guess} verifying the assumed form of $f()$ though this is probably not its most general form. Doing the same thing for a small circuit $C_2$ anticlockwise round $z_2$ gives the equivalent result $f(z)=f(z)\exp(\frac{4\pi^2}{\ln(t)+2n\pi i})$. \section{\label{errors_here}Simplest solutions of the equations defining singular points} ******* This section seems as if there are some very important results to be found but it needs quite a lot of work yet ********** *'s indicate likely theorems that have not yet been proved. Let the binary relation $\to$ on analytic functions be defined by\newline $f()\to g() \Leftrightarrow$ there exists an analytic function $h()$ such that $f()=h(g())$. Then the relation $\to$ that points towards the simpler function is reflexive and transitive. Also \begin{Theorem}\label{thm10.1} If $f()\to g()$ and $g()\to f()$ then $f(z)=\frac{a+bg(z)}{c+dg(z)}$ for some finite constants $a,b,c,d\in\mathbb{C}$.\end{Theorem} \begin{proof} Suppose $f()\to g()$ and $g()\to f()$ then $f()=h_1(g())$ and $g()=h_2(f())$ for some analytic functions $h_1()$ and $h_2()$, and therefore $f()=h_1(h_2(f()))$ i.e. $h_1(h_2())= I()$ which has no singular point. By Theorem~\ref{nsp} $h_1()$ can have no singular point and is therefore a bilinear function and so $f(z)=\frac{a+bg(z)}{c+dg(z)}$. \end{proof} Suppose a set $S$ of analytic functions is such that if $f()\in S$ then $h(f())\in S$. Then this set is determined by the set $R\subseteq S$, the root functions, such that for any analytic function $f()$ in $S$ there exists a member $g()\in R$ such that $f()\to g()$. Such a set will be called a rooted set. Suppose a single root function $r()$ acts a root for $S$ i.e. $\forall f()\in S[f()\to r()]$. Suppose another function $r_1()$ also has this property, then $\forall f()\in S[f()\to r_1()]$ and in particular $r()\to r_1()$. Likewise $r_1()\to r()$. Then by Theorem~\ref{thm10.1} $r(z)=\frac{a+br_1(z)}{c+dr_1(z)}$. Such a rooted set will be called singly-rooted. Thus the root functions associated with a singly rooted set are related by a bilinear transformation. From Theorem~\ref{thm10.1} any root function $k()$ is unique up to a bilinear function or transformation (also known as a M\"obius transformation or a linear fractional transformation) i.e $k_1(z)=\frac{a +bk(z)}{c+dk(z)}$ so a root function is actually a set of functions each member of which is related to any other member like this for some set of values $a,b,c,d\in \mathbb{C}$ such that $ad-bc\ne 0$. The terminology below will for simplicity refer to this special set just as a single function, the root function. \begin{comment} These comments relate to earlier work which used a circular argument showing that $\oplus$ always exists which is false. An immediate consequence of this is that the multiple intersection of a set of singly-rooted sets is a singly rooted set. Also the binary operation that gives the root function $f()\oplus g()$ of a pair of singly-rooted sets with root functions $f()$ and $g()$ is both commutative [$f()\oplus g()=g()\oplus f()$] and associative [$(f()\oplus g())\oplus h()=f()\oplus (g()\oplus h())$], and $f()\oplus f()=f()$. The symbol $\oplus$ was chosen because the operation has some properties of $+$ and is related to composition which is denoted by $o$. \end{comment} \begin{Lemma}Every analytic function is in the set rooted by a left-unique function. \end{Lemma} \begin{proof}If $g()$ is left-unique then $f(z)=f(g^{o-1}(g(z)))$ so $f()\to g()$.\end{proof} \begin{Theorem} If $gof()\equiv f(g())=g(f())$ and $g()$ is left-unique and right-unique then $f()\oplus g()=gof()$. \end{Theorem} \begin{proof} The condition on $g()$ gives $g^{o-1}(g())=I$ and $g(g^{o-1}())=I$ and also $f(g())=g(f())$. Suppose $l(z)=h_1(f(z))$ and $l(z)=h_2(g(z))$ for some arbitrary analytic functions $h_1()$ and $h_2()$. Then $f(z)=f(g^{o-1}(g(z)))=g^{o-1}(g(f(z)))=g^{o-1}(f(g(z)))$. Therefore $f(g^{o-1}(w))=g^{o-1}(f(w))$ generally [where $w=g(z)$] and $l(z)=h_1(f(g^{o-1}(g(z)))=h_1(g^{o-1}(f(g(z)))=h_3(f(g(z))$ where $h_3()=h_1(g^{o-1}())$. According to the criterion for a root function, $f(g())$ is the required root function for the set of possible analytic functions $l()$. \end{proof} \begin{comment} \begin{itemize} \item Derive any other properties $\oplus$ has in relation to $o$. \item Derive an equivalent definition of $\oplus$ depending only on logic as would happen for relations over an arbitrary set instead of $\overline{\mathbb{C}}$. \item find some other examples of $\oplus$ that can be solved explicitly. \end{itemize} \end{comment} This shows that the general solution is determined by one particular solution. Could $f(z)+g(z)$ be the root function? If so, then $f(z)+g(z)=h_1(f(z))$ for some function $h_1()$. Then $h_1(z)=z+g(f^{o-1}(z))$ formally, and this requires $f^{o-1}(f(z))=z$ i.e. $f()$ is left-unique. Likewise for $f(z)+g(z)=h_2(g(z))$ showing that this also requires $g()$ to be left-unique. This shows that $f()+g()\in S$ if $f()$ and $g()$ are left-unique. $f()+g()$ is to be the root function also requires that for any function $l()$, $l()\to f()\text{ and }l()\to g()\Rightarrow l()\to f()+g()$. \begin{comment} \begin{Theorem} If $f()$ and $g()$ are left-unique and right-unique analytic functions then so is $f()+g()$. \end{Theorem} \begin{proof} If $f()$ and $g()$ are left-unique and right-unique analytic functions then by Lemma~\ref{lemma6.5} $f()$ and $g()$ have no singular point and by **** they are both bilinear functions i.e. $f(z)=\frac{a_1+b_1z}{c_1+d_1z}$ $g(z)=\frac{a_2+b_2z}{c_2+d_2z}$ If $f()$ and $g()$ are left-unique then $\forall z_1,z_2\in\overline{\mathbb{C}}[f(z_1)=f(z_2)\Rightarrow z_1=z_2]$ and $\forall z_1,z_2\in\overline{\mathbb{C}}[g(z_1)=g(z_2)\Rightarrow z_1=z_2]$. Also suppose contrary to the theorem that $f()+g()$ is not left-unique. Then $\exists z_1,z_2\in\overline{\mathbb{C}}[f(z_1)+g(z_1)=f(z_2)+g(z_2)\text{ and }z_1\ne z_2]$. That is $(f()+g())(z_1)=(f()+g())(l(z_1))$ i.e. $f()+g()$ satisfies an equation of the type \eqref{eq3} for some $z_1,z_2=l(z_1)\in\overline{\mathbb{C}}$ with $l(z_1)\ne z_1$ i.e. $f()+g()$ is not locally left-unique at $z_1,f(z_1)+g(z_1))$ and $z_2,f(z_1)+g(z_1))$. Because $f()$ and $g()$ are continuous as $z_1$ varies, so will $l(z_1)$ be, and in fact the function $l()$ must also be analytic *. The solution of $z_1=l(z_1)$ will exist by Theorem~\ref{thm5.4} (even if $l(z)=z+c$ where $c$ is a constant it is $\infty$) say $z_3$ and be a point in $\overline{\mathbb{C}}$ where $f()+g()$ has a singular point and $f()+g()$ will be not locally left-unique there. Therefore either $f()$ or $g()$ must also have a singular point at $z_3$ of this type *** and this is inconsistent with the premises of the theorem. \end{proof} \end{comment} Suppose now that $f()$ is not left-unique for example $f(z)=l(z)^2$ then the problem is to find the intersection $h_1(l()^2)\cap h_2(g())$. A common type of equation defining behaviour around a singular point is \begin{equation}\label{1}f(z)=g_1(f(g_2(z))\end{equation} where $g_1()$ and $g_2()$ are right-unique functions. The more general form \begin{equation}\label{eq2}f(z)=g_1(z,f(g_2(z)))\end{equation} occurs later. Most of the examples above are actually special cases of \begin{equation}\label{eq3}f(z)=f(g_2(z))\end{equation} [could this be generalised to $f(z)=f(z,g_2(z))$?] or \begin{equation}\label{eq9}f(z)=g_1(z,f(z))\end{equation} which are themselves special cases of \eqref{eq2}. Equation \eqref{eq3} can be iterated to give $\label{eq_it}\forall n\in\mathbb{N}[f(z)=f(g_2^{on}(z))]$ which is equivalent to \eqref{eq3}. This can be expressed as \begin{equation}\label{eq_it_alt}\exists n\in\mathbb{N}[z_2=g^{on}_2(z_1)]\Rightarrow f(z_1)=f(z_2). \end{equation} If $l()$ is any function then from \eqref{eq_it_alt} it follows that $\exists n\in\mathbb{N}[z_2=g^{on}_2(z_1)]\Rightarrow l(f(z_1))=l(f(z_2))$ i.e. if $f()$ is a solution of \eqref{eq3} so is $l(f())$. If $l()$ is analytic so will be $l(f())$. Let $f_s()$ be a special solution of \eqref{eq3} that satisfies in addition the converse of \eqref{eq_it_alt} i.e. \begin{equation}\label{simp_1}f_s(z_1)=f_s(z_2)\Rightarrow \exists n\in\mathbb{N}[z_2=g^{on}_2(z_1)].\end{equation} This introduces the equivalence relation $\sim$ defined by $z_1\sim z_2\Leftrightarrow \exists n\in\mathbb{N}[z_2=g^{on}_2(z_1)\text{ or }z_1=g^{on}_2(z_2)]$ and states that the values of $f_s()$ are in one to one correspondence with the equivalence classes of $\sim $. Any solution $f(z)$ of \eqref{eq3} is a function of the equivalence classes i.e. its value is the same for each member of the same equivalence class, (but regarded as a function of the equivalence classes, is not necessarily left-unique), therefore it can be written as a function of a $f_s()$ i.e. $f(z)=h(f_s(z))$. Until now $f()$ and $f_s()$ were tacitly assumed to be right-unique but this does not have to be the case because pairs or sets of values of $f_s()$ will then be in left-unique correspondence with the equivalence classes of $\sim$. Because the functions $f()$ and $f_s()$ are analytic, $h()$ will be also. This works if $f_s()$ exists. The extension to multiple simultaneous equations of type \eqref{eq3} will also depend on the analogous existence theorem. Another approach is to replace the equations by the limit of some equations for algebraic functions for which the uniqueness of the solution given all the singular behaviours at each of the singular points has been established. Because of lemma~\ref{lemma1}, every solution of \eqref{eq3} has a singular point at points $z$ where $z=g_2(z)$ for example $(f_s(z))^{-2}$. This example also has a singular point where $f_s(z)=0$ and suggests that of all the analytic solutions of \eqref{eq3}, the special solutions $f_s()$ that also satisfy \eqref{simp_1} have singular points only where $z=g_2(z)$. To prove this suppose $z_1\ne g_2(z_1)$. The condition for $f_s()$ to have no singular point at $P$, $(z_1,f_s(z_1))$, using \eqref{eq_it_alt} and \eqref{simp_1}, is that there is a neighbourhood $N$ of $P$ such that for all points $(z_2,f_s(z_2))$ and $(z_3,f_s(z_3))\in N$, $z_2=z_3\Leftrightarrow z_2\sim z_3$. The last condition reduces to $\exists n\in\mathbb{N}[z_2=g_2^{on}(z_3))]\Rightarrow z_2=z_3$. To establish this it is sufficient to choose $N$ so small that if $z_3$ is included by being sufficently close to $z_1$ that none of $g_2(z_3), g_2(g_2(z_3))$ etc. are included i.e. the image of $N$ under $g_2()$ must not overlap $N$ itself. There is another case i.e. $g_2^{ok}(z_1)=z_1$ where $k\ge2$ where the proposition is also true. Therefore these special fundamental solutions of \eqref{eq3} that also satisfy \eqref{simp_1} will be called the {\em simplest solutions} of \eqref{eq3}. Let $f_s^*()$ be another function that satisfies the conditions on $f_s()$ above then $f_s^*(z)=h^*(f_s(z))$. If $h^*()$ is any function without a singular or an inversion point (i.e. a linear function) then by Lemma~\eqref{lemma3} $f_s^*()$ will satisfy this condition i.e. the set of {\em simplest solutions} of \eqref{eq3} must include $a+bf_s(z)$ if $f_s(z)$ is included. Can there be any more? Any other such solution must take this form with a different function $h^*()$ that will be nonlinear and must have at least two singular or inversion points somewhere and $h^*(a+bf_s(z))$ must have no singular point except when $z$ satisfies $z=g_2(z)$ for all $a,b\in\overline{\mathbb{C}}$. This is impossible because the argument of $h^*()$ can then take any value, so this proves \begin{Theorem} The set of simplest solutions to \eqref{eq3} i.e. those that also satisfy \eqref{simp_1} have singular points only where $z=g_2(z)$ where $g_2()$ is as in \eqref{eq3}. This set is the set $a+bf_s(z)$ for arbitrary $a,b\in \overline{\mathbb{C}}$ if $f_s(z)$ is itself a simplest solution of \eqref{eq3}. Any solution to \eqref{eq3} can be written as $h(f_s(z))$ for some simplest solution $f_s(z)$ for some analytic function $h()$. \end{Theorem} %$\exp(\ln(z))=z$ after ``plugging" the removable singularity at 0 but $\ln(\exp(z))=z+2n\pi i\text{ for all }n\in \mathbb{Z}$. It is interesting to note that in the examples the index set $\mathbb{N}$ can sometimes be replaced by a finite set. If $g_2()$ is not a linear function the equation $z=g_2(z)$ that determines the singular point could have many solutions, and $g_2()$ itself could be described by another equation of the type \eqref{eq3} or \eqref{eq9} etc.. In such a case the original equation \eqref{eq3} together with other similar equations to determine $g_2()$ could determine behaviour at a set of singular points simultaneously. In such a case it might be a good idea to try to solve for the singular points and then with $g_2()$ replaced by linear functions that give the same singular points, analyse each separately using the results in Section \ref{sec9} and then try to reconstruct the original function but note example \eqref{nonlin} indicating that in this case an infinite number of singular points can sometimes occur. A similar argument to that applied to \eqref{eq3} can be applied to \eqref{eq9} giving the iteration as \begin{equation}\label{eq_it2}\forall n\in\mathbb{N}[f(z)=g_1^{on}(z,f(z))]\end{equation} which is equivalent to \eqref{eq9} where $g_1()$ appears $n$ times in $g_1^{on}(z,f(z))=g_1(z,g_1(z,g_1(z,\ldots g_1(z,f(z)\ldots)$. ********************The {\em simplest solution} satisfies \begin{equation}\label{simp_2}z_1=z_2\Rightarrow\exists n\in\mathbb{N}[f(z_1)=g_1^{on}(z_2,f(z_2))]\end{equation} where the justification is similar i.e. \eqref{eq_it2} gives all the values of the function $f(z)$ that are {\em determined} by one value of $f(z)$ whereas \eqref{simp_2} states that for any value of $f(z)$ at the same point $z$, it must be one of the values in \eqref{eq_it2}. Now consider iteration applied to \eqref{eq2} which gives \begin{equation}\begin{array}{l}f(z)=g_1(z,g_1(g_2(z),f(g_2^{o2}(z))))=...=\\ g_1(z,g_1(g_2(z),g_1(g_2^{o2}(z),g_1(g_2^{o3}(z),g_1(g_2^{o4},\ldots)\ldots)= [g_1((),f(g_2())]^{on}(z)\end{array}\end{equation} where $g_1$ appears $n$ times in this expression. Now suppose $g_2^{on}()$ is the identity function $:z\to z$ then \begin{equation}\label{eq2_it} f(z)=g_1(z,g_1(g_2(z),g_1(g_2^{o2}(z),\ldots g_1(g_2^{o(n-1)}(z),f(z))\ldots).\end{equation} This is last expression depends independently on $z$ and $f(z)$ through the functions $g_1()$ and $g_2()$ and can therefore be written as $k(z,f(z))$ i.e. \eqref{eq2_it} can be written in the form \eqref{eq9} for different $g_1()$ and $g_2()$. Also it is concievable that \eqref{eq2_it} for some value of $n$ takes the simpler form \eqref{eq3} again for different $g_1()$ and $g_2()$. In either of these cases the {\em simplest solution} of the respective iterated form of \eqref{eq2} can be defined as above. If this can be done for both cases the following example suggests this might define the {\em simplest solution} for \eqref{eq2} itself. There are many results that can be obtained relating the solution sets of \eqref{1} with different values of $g_1()$ and $g_2()$. If \eqref{1} holds then the same relationship holds with $f()$ replaced by $k(f(l()))$, $g_1()$ replaced by $k(g_1(k^{o-1}()))$ and $g_2()$ replaced by $l^{o-1}(g_2(l()))$. Making these substitutions gives the same relationship with the function $k()$ applied to both sides and expressed in terms of the independent variable $w$ given by $z=l(w)$. For example suppose $k(z)=az+b$ and $l(z)=cz+d$ then the function $f^*(z)=k(f(l(z)))=af(cz+d)+b$ satisfies $f^*(z)=g^*_1(f^*(g^*_2(z)))$ i.e. \eqref{1} with $g^*_1(z)=ag_1((z-b)/a)+b$ and $g^*_2(z)=(g_2(cz+d)-d)/c$. If in equation~\eqref{1} $g_1^{o-1}()$ is applied to both sides and the result expressed in terms of the variable $w=g_2(z)$ then the same relationship holds with $g_1()$ replaced by $g_1^{o-1}()$ and $g_2()$ replaced by $g_2^{o-1}()$. The inverse functions of both sides of Equation~\eqref{1} again give an equation of the same form showing that $f^{o-1}$ satisfies the equation of the same form but with $g_1()$ replaced by $g_2^{o-1}()$ and $g_2()$ replaced by $g_1^{o-1}()$. In these general arguments, it has to be borne in mind that $f^{o-1}(f(z))$ could have several components and is not necessarily just the identity function as in section \ref{sec1}. \section{Further thoughts on solutions to equations \eqref{1} and \eqref{2}} Suppose a general solution of \eqref{2} is of the form \begin{equation}\label{2_sol}f(z)=F(h(G(z))\end{equation} for some fixed functions $F$ and $G$ where $h()$ is an arbitrary analytic function. Now suppose in addition to \eqref{2} that $f()$ has no singular points other than those required by \eqref{2} including no other conditions that could modify behaviour at the singular point(s) required by \eqref{2} i.e. at $(z,f)$ such that \begin{equation}\label{sing2}\begin{array}{l}z=g_2(z)\\f=g_1(z,f)\end{array}.\end{equation} Next consider the finite singular points of $h()$. Let $z=s$ be such a point then by \eqref{2_sol} any point $z$ such that $G(z)=s$ will be a singular point of $f()$. By assumption this cannot happen [unless this $z$ with a value for $f$ satisfies \eqref{sing2}]. An additional condition on $h()$ at the singular points would correspond to an additional condition on $f()$ which is assumed not to happen therefore $h()$ has no finite singular points and therefore $h(z)=a+bz$ and $f(z)=F(a+bG(z))$. This proves that \begin{Theorem} If the general solution of \eqref{2} is \eqref{2_sol} and if $f()$ has no singular points other than those required by \eqref{2} including no other conditions that could modify behaviour at the singular point(s) required by \eqref{2} then $f(z)=F(a+bG(z))$. \end{Theorem} Consider examples where more than one singular point is analysed like this in detail and then I expect \eqref{1} or \eqref{2} will only asymptotically hold close to the corresponding singular point. Consider for this a linear combination (LC) of the minimal solutions for each separate singular point. By minimal I mean solutions that have no other finite singular points (see below). This LC will have precisely the asymptotically defined behaviours at the singular points because all the other terms will not have a singular point at each of them. This I think can be generalised to nonlinear combinations, if the condition of minimality is dropped. i.e. find the general solution of a set of simultaneous asymptotically defined relations about a set of singular points as some arbitrary analytic function of basic solutions to them singly? Questions about future research \begin{itemize} \item Is it possible and practical to use the closure properties to prove statements by a kind of induction i.e. if a statement is true for the constant function, and it is true for functions $f$ and $g$ implies it is true for their union, sum, product, $f'()$ (the derivative of $f()$), $\int_0^{x}f(s)ds$,composition, $f^{o-1}$(inverse of $f$)., etc. then it is true for the whole algebra of functions? \item Does the algebra include the solutions of differential equations? eg $f''(x)+k(x)f(x)=0$ Perhaps this needs the limit of a sequence of functions to be included if the sequence is already included because any differential equation can be written as an integral equation which can be solved as the limit of an iteration. Then I think differentiation and integration can be removed from the closure operations. \item Can algebraic functions be characterised as those satisfying \eqref{eq2} with linear $g_1()$ and $g_2()$? \item What role is played by the operation of getting the {\em simplest solution} of \eqref{eq2} with given $g_1()$ and $g_2()$? How should this be done in general? \item Consider the binary operation which is the {\em simplest common solution} $f()$ of $f(z)=h_1(f_1(z))$ and $f(z)=h_2(f_2(z))$ for fixed $f_1()$ and $f_2()$, but arbitrary $h_1()$ and $h_2()$. \item What about the solutions of different equations say $f(z)=g_1(f(f(z)))$? \end{itemize} \begin{thebibliography}{99} \bibitem[Churchill et al.]{CBV}Churchill R.V., Brown J.W., Verhey R.F., Complex Variables and Applications, Third Edition, McGraw-Hill Kogakusha 1974 \bibitem[Nixon2013]{jhn2013}Nixon J., Theory of algebraic functions on the Riemann Sphere Mathematica Aeterna Vol. 3, 2013, no. 2, 83-101\newline \href{https://www.longdom.org/articles/theory-of-algebraic-functions-on-the-riemann-sphere.pdf}{https://www.longdom.org/articles/theory-of-algebraic-functions-on-the-riemann-sphere.pdf} \bibitem[Encyclopedia of Mathematics]{eom}*\newline \href{https://encyclopediaofmath.org/wiki/analytic_function}{https://encylopediaofmath.org/wiki/Analytic-function} \end{thebibliography} \end{document}