Find a holomorphic function $f : \Bbb C^* \to \Bbb C$ such that $$\int_0^{2 \pi}f(e^{it})e^{int} dt = \frac{2 \pi}{\lvert n\rvert !} \quad \forall n \in \Bbb Z $$
I said that $$\int_0^{2 \pi}f(e^{it})e^{int} dt = \int_0^{2 \pi}f(e^{it})e^{i(n-1)t}\frac{ie^{it}}{i} dt = \frac{2 \pi}{\lvert n\rvert !} $$ implies that $$\frac{1}{2 \pi i} \int_{\partial\Bbb D(0,1)} f(z)z^{n-1} dz = \frac1{\lvert n\rvert !}= res_0(f(z)z^{n-1})$$
So we have that $f(z)z^{n-1}=\cdots + \frac1{\lvert n\rvert !z} + \cdots \implies f(z)=\cdots + \frac1{\lvert n\rvert !z^n} + \cdots$. Since this is true for all integer, the Laurent series of $f$ is given by $$f(z)= \sum_{-\infty}^{\infty}\frac{1}{\lvert n \rvert !}z^n = e^z+ e^{1/z}$$.
Is it correct ?