Prove that there are unique entire functions $A_n$ such that for all $\forall (w,z) \ \mathbb{C} \times (\mathbb{C} $\ 0) fulfilling the equation $\exp(\frac{w}{2}(z-\frac{1}{z}))=\sum_{n=-\infty}^{\infty} A_n(w) z^n$
what I have tried: I have tried to separate the exponent of the exponential function and bringing the coefficients into the desired form.
$\exp(\frac{w}{2}z)=\sum_{n=0}^{\infty} \frac{(\frac{w}{2})^n}{n!} z^n = \sum_{n=0}^{\infty} A_{n_1}(w) z^n $
$\exp(-\frac{w}{2}\frac{1}{z})=\sum_{n=0}^{\infty} \frac{(-\frac{w}{2})^n}{n!} z^{-n} = \sum_{n=0}^{\infty} A_{n_2}(w) z^{-n} $
so $A_{n_1}(w) = \frac{(\frac{w}{2})^n}{n!} $ $A_{n_2}(w) = \frac{(-\frac{w}{2})^n}{n!} $
so I do not know whether this is helpful, but I think there must be a manipulation trick to merge the functions and series together.
If you fix $w \ne 0$ the function $\exp(\frac{w}{2}(z-\frac{1}{z}))$ has obviously a Laurent expansion $\sum_{n \in \mathbb Z} c_nz^n$ (uniformly convergent on compacts in the punctured plane) with coefficients depending on $w$ which you can call $A_n(w)$.
Clearly $A_n(0)=0, n \ne 0, A_0(0)=1$ since $\exp(\frac{w}{2}(z-\frac{1}{z}))=1, w=0$ so the only thing to prove is that those are differentiable and that is straightforward as (integrating term by term on the unit circle which is allowed by uniform convergence) $$2\pi i A_k(w)=\int_{|z|=1}\exp(\frac{w}{2}(z-\frac{1}{z}))z^{-k-1}dz$$ gives $A_k(w)$ unique and entire.