As is said in the title I would like to evaluate the residue of $f(z) = e^{\frac{1}{1+√z}}dz$ in the point $z = 1$ for all branches of $e^{\frac{1}{1+√z}}$.
This is new for me, but after reading some exercises I believe the method is to find the points $(1,w)$ on Riemann surface, for each of them find a local coordinate (or parameter? I'm not sure there's a difference), and then express our initial function in terms of that corrdinate.
If this is true and enough, I guess that here the points of that Riemann surface are $z_1 = (1,e^{\frac{1}{2}})$ and $(1,\infty)$. (If the exponential function has only one branch, which I'm not sure, since its inverse function has 2 branches).
I believe that $z$ is a local coordinate for both $z_1$ and $z_2$, by Theorem on implicit function. But how can we expand $f(z)$ with respect to $z$?
I believe this kind of exercise can be done quite quickly once we get the method, hence I hope you can help me understand it so I can do it again !
Edit
I may have found another way to evaluate this residue but I'm not sure either, so here is what I did:
I noted that $z$ is a pole of order 1 for $f(z)$ in the branch defined by $\sqrt{1} = - 1$, meaning that in this branch its residue could befound using the following formula:
$$res_{z = z_0}f(z) = \lim\limits_{z\to z_0}(z - z_0)f(z)$$ which equals $0$ here.
So this would mean that $res_{z = 1}f(z) = 0$ in the branch defined by $\sqrt{1} = - 1$ and just doesn't exist in the other branch. Is this correct?
@Gary ?
Thanks