I'm trying to solve the following problem:
Suppose we know that everywhere outside the circle $C_R$, radius $R$ centered at the origin, $f(z),g(z)$ are analytic with $\lim_{z \to \infty} f(z) = C_1$ and $\lim_{z \to \infty} zg(z) = C_2$ where $C_i$ are constant. Show: $$\frac{1}{2\pi i} \oint_{C_R}g(z)e^{f(z)} = C_2e^{C_1}$$
I know this problem involves residue at infinity. But I have issues trying to calculate it. By the well-known formula we get:
$$\frac{1}{2\pi i} \oint_{C_R}g(z)e^{f(z)}dz = -Res_{z=\infty}g(z)e^f(z) = Res_{z=0}\frac{g(1/z)e^f(1/z)}{z^2}$$ since the residue at infinity of $f$ is defined by
$$Res_{z=\infty}f(z) = - Res_{z=0}\frac{1}{z^2}f(1/z)$$
So I'm basically having issues calculating said residue.