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Let $a,b,c$ be real variables.

Let $z$ be a complex number and $g(z) = exp(z)(z-a)(z-b)(z-c)$.

Let $f(z)$ be the functional inverse of $g(z)$ such that $f(g(z)) = g(f(z)) = z$. Now $f(z)$ must have singularities where $g'(z) = 0$ hence since this equation has at most 3 distinct complex solutions and $g(z)$ is an entire function ; $f(z)$ has at most 3 distinct singularities.

Lets call those singularities $A,B,C$.

Now I wonder about Puiseux series

What are the Puiseux series at the points $A,B,C$ ?

I note that although not written in the link , a Puiseux series power series part is suppose to have nonzero radius of convergence.

Depending (for instance) on how the radius of convergence is for the power series part , there are at most 4 possible Puiseux series for point $A$ because there are 3 singularities. ( those being valid everywhere , valid close too from $A$ to $B$, valid close too from $A$ to $B$ and $C$, valid close too from $A$ to $C$)

In fact the amount of Puiseux series up to translation with $m$ singularities is at most $2^m - 1$. And that makes $7$ in this case.

So what are the Puiseux series for the inverse of $f(z)$ expanded at the singularities?

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$f$ has a singularity where $g'(f)=0$, not just where $g'=0$. – WimC Nov 9 '12 at 19:48
@WimC I do not know what you mean. Note that essential singularities must occur at every branch since they are unbounded. – mick Nov 9 '12 at 20:03
Consider a simpler example $g(z)=1+z^2$. Where does its inverse have singularities? – WimC Nov 10 '12 at 6:25
$g'(z) = 0$if $z=0$ so $g(0)$ is a singularity of the inverse of $g$. – mick Nov 10 '12 at 16:03
That's not what you stated in (or at least what I read) your question. "Now $f(z)$ must have singularities where $g′(z)=0$". In this example I'd interpret that as $f$ has a singularity at $z=0$ since $g'(0) = 0$. – WimC Nov 11 '12 at 8:29

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