# Ahlfors method for partial fraction decomposition

On page 31 in Complex Analysis by Ahlfors, he discusses decomposing a rational function into $$R(z) = G(z) + H(z)\tag{12}$$ where $G(z)$ is a polynomial without a constant term and is the singular part. Then he goes on to say let $\beta_i$ be the distinct finite poles of $R(z)$. The function $R(\beta_i + 1/\zeta)$ is a rational function of $\zeta$ with a pole at $\infty$. By use of the decomposition of $(12)$, we can write $$R(\beta_i+1/\zeta) = G_i(\zeta) + H_i(\zeta)$$ or with a change of variables $$R(z) = G_i(1/(z-\beta_i)) + H_i(1/(z-\beta_i))$$ Using the method developed in the text, write $$\frac{z^4}{z^3-1}$$ in partial fractions.

If I just do division, I get $$\frac{z^4}{z^3-1} = z + \frac{z}{z^3-1}$$ and the solution is quick and efficient.

If I try to use the method developed in the text, (1) I am not able to get the correct answer and (2) it seems to hold no advantage to polynomial division.

The poles are $z=1, e^{2\pi i/3}, e^{4\pi i/3}$ so $$R(1+1/\zeta) = \frac{(1+1/\zeta)^4}{(1+1/\zeta)^3 - 1}$$ Should I expand the polynomials or multiple through by $\zeta^4$ and then expand? Both methods for the first pole would take more time then just polynomial division so I am under the impression there has to be an option three or this method wouldn't be of benefit to learn.

You would write $R(\beta_i + 1/\zeta)$ as a quotient of polynomials, in your example, multiply numerator and denominator with $\zeta^4$ to obtain
$$R(1 + 1/\zeta) = \frac{(\zeta+1)^4}{\zeta\bigl((\zeta+1)^3 - \zeta^3\bigr)},$$