I have been trying to evaluate the residue at infinity of the following function:
$\exp(-\frac{1}{3}Log(1-z))\exp(-\frac{2}{3}Log(1+z))$
The first Log (multiplied by $-\frac{1}{3}$) has branch cut $[0,\infty)$ and the second has branch cut $(-\infty,0]$, i.e the principal branch.
Therefore the left term has branch cut at $(-\infty ,1]$ and the right term has branch cut at $(-\infty,-1]$.
Therefore the product of those two terms has branch cut $[-1,1]$ and one can show this by determining the continuity of that product over the interval where both branch cuts of the individual terms coalesce.
I successfully evaluated the residue at infinity of that function by evaluating the integral of that product over an arbitrarily large circle centre $0$, and got $-exp(-\pi/3)$ which I believe is correct, integrating clockwise over the circle.
My question is, that in order to evaluate this integral over this arbitrarily large circle, I made the substitution $Log(1-R\exp(i \theta)) \approx Log(-R\exp( i \theta)$ and $Log(1 + R\exp(i \theta)) \approx Log(R\exp(i \theta))$ which I thought to be true (with respect to the individual branch cuts of each Log) to a degree of error that would produce an overall error going to $0$ of the integral I wanted to evaluate.
Now I would like to formalise these replacements, and I think perhaps one way to do so would be to find Laurent series with arbitrarily large exterior radii for the product, but I'm not sure how to do so at all, without just calculating the coefficients by hand term by term but that seems like a painful process… and I'm not even sure if this is a possible approach or if there is an easier justification for these replacements
Any help would be greatly appreciated!
Best Answer
As a multi-valued function, $f(z) = (1 - z)^{-1/3} (1 + z)^{-2/3}$.
$z = \infty$ is not a branch point, so there are three branches which are regular at infinity. With the principal value of the power function, they can be written analytically as $$\frac 1 {e^{i (\pi + 2 \pi k)/3} z \left( 1 - \frac 1 z \right)^{1/3} \left( 1 + \frac 1 z \right)^{2/3}}.$$ The residue at infinity is $-e^{-i (\pi + 2 \pi k)/3}$. Your choice of the logarithms corresponds to $k = 0$.