Maclaurin series for $ f(f(z)) = \exp(z) - \exp(-z) + \exp(-z/2) - \exp(-z/5)+ \exp(-z/6) - \exp(-z/9)$

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I wonder about the Maclaurin series of the analytic $f(z)$ such that

$$ f(f(z)) = g(z) = \exp(z) - \exp(-z) + \exp(-z/2) - \exp(-z/5)+ \exp(-z/6) - \exp(-z/9) $$

Since zero is a fixpoint and $g'(0) = \frac{74}{45} > 1$ we can find such a Maclaurin series of $f(z)$ in many ways.

Using Koenigs function for instance. Or by using Carleman matrices.

We then can compute

$$f(z) = a_1 z + a_2 z^2 + a_3 z^3 + ...$$

I would like many terms of these coefficients and I wonder how many terms we need before one of them is negative ?

I also wonder about the exact radius of $f(z)$. Ofcourse the radius is equal or less than the distance to the next fixpoint of $g(z)$.

I know there is a fixpoint for $g$ around $-0.218 + 1.504 i$ and its complex conjugate.

Is the radius equal to the closest distance to a point $A$ where $f(f(A))=g(A)$ still holds ?