I'm trying to understand a simple example of the circle method to solve a trivial problem: given $k \in \mathbb{N}$, determine the number of possible representations of n $\in$ N as a sum of exactly $k$ natural numbers.
Q1: I cannot see how obviously A is derived as the expansion of $z^n$ will not result in $(1-z)^{-1}$?
Q2: How is Cauchy's theorem applied to get B?
(A) You certainly already know the series of equalities: $$ \begin{aligned} 1 &=\frac{1-z}{1-z}\ ,\\ 1+z &=\frac{1-z^2}{1-z}\ ,\\ 1+z+z^2 &=\frac{1-z^3}{1-z}\ ,\\ 1+z+z^2 +z^3 &=\frac{1-z^4}{1-z}\ ,\\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ &\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \vdots\ \\ 1+z+z^2 +z^3 + \dots+ z^N&=\frac{1-z^{N+1}}{1-z}\ , \end{aligned} $$ and now we need $|z|<1$ to be able to pass to the limit on the right side. This was (A).
For (B), let us rewrite: $$ \frac 1{(1-z)^k\ z^{n+1}} = \Big(\ 1+z+z^2 +z^3 + \dots+ z^n+\dots\ \Big)^k\cdot \frac 1{z^{n+1}}\ . $$ After we take the integral on some small contour around $0$, up to the $2\pi i$ factor, we pick the coefficient of $x^n$ in the expansion $$ \Big(\ 1+z+z^2 +z^3 + \dots+ z^n+\dots\ \Big)^k\ , $$ equivalently, in the polynomial $$ \Big(\ 1+z+z^2 +z^3 + \dots+ z^n\ \Big)^k\ . $$ This polynomial is a product of $k$ factors, we take
each monomial from the first factor, $z^{s_1}$, "with" (i.e. multiplied by)
each monomial from the second factor, $z^{s_2}$, "with"
and so on, "with"
each monomial from the last factor, $z^{s_k}$,
and we want to get $z^n$, so $$ \begin{aligned} &\text{Coefficient of $z^n$ in } \Big(\ 1+z+z^2 +z^3 + \dots+ z^n\ \Big)^k \\ &\qquad =\sum_{ \substack{0\le s_1\le n\\0\le s_2\le n\\\vdots\\0\le s_n\le n\\s_1+s_2+\dots+s_k=n} }1 \\ &\qquad = r_k(n) \ . \end{aligned} $$