Laurent Series Expansion Logic

Question

Relative to the below image, I am curious about the progression from equation 3.2 to equation 3.3, then from equation 3.3 to equation 3.4.

I understand the logic in 3.2. I understand that a Laurent expansion about a complex function $f(z)$ is $$f(z)=\sum_{n=-\infty}^{\infty}a_{n}(z-c)^{n}.$$

$a_{n}$ is a function of a line integral

$$a_{n}=\frac{1}{2\pi i}\oint_{\gamma}^{}\frac{f(z)dz}{(z-c)^{n+1}}.$$

I can see that $c=0$, but I don't understand what $A_{n}$ represents and thus do not understand how the writer progressed from 3.3 to 3.4 either.

Thank you!

Answer 1

From $\varphi(z)=\sum_{n=-\infty}^\infty A_n(q)z^n$ we see that the coefficient of $z^n$ is equal to $A_n(q)$. From $\varphi(z)=\sum_{n=-\infty}^\infty A_n(q)z^{n+1}q^{n+1}$ we see that the coefficient of $z^n$ is $q^nA_{n-1}(q)$. Thus, $A_n(q)=q^nA_{n-1}(q)$.

Answer 2

From the definition of $\phi\left(z\right)$ as the left hand side of 3.1 it is clear that it may be written as a Laurent series about $z=0$; to see this think of multiplying out the brackets; the products of $z^{-1}q^{n-1}$ with itself produce negative powers of $z$.

Next consider an annulus in which this Laurent series converges (a deleted neighbourhood of $z=0$ will do the trick since $z=0$ is an isolated singularity of $\phi\left(z\right)$). Laurent's theorem says then that there exist constants $A_n$ as per the first equality in equation 3.3. The second equality in 3.3 follows from the conclusion of 3.2 that $\phi\left(zq\right)=z^{-1}q^{-1}\phi\left(z\right)$. The third equality in 3.3 follows from the first, by substituting $\phi\left(zq\right)=\sum_{n=-\infty}^{+\infty}A_n\left(q\right)(zq)^n$ into $zq\phi\left(zq\right)$. Then we get 3.4 by equating coefficients.

The existence of the $A_n$ is guaranteed by Laurent's theorem, and at 3.3 all we know is that they exist (and depend on $q$); then we proceed to calculate them via the recurrence relation 3.4.