Trouble with understanding the proof of the contribution of the principle part on the coefficients of the function

Question

• $PP(f;z_0)$ represents the principle part of $f(z)$ at pole $z_0$

• $r$ represents the order of the pole

My confusion arises in the third row where the paper gets that $$\left(1-\frac{z}{z_0}\right)^{-j}=\sum_{n≥0} {{n+j-1}\choose{n}}\left(\frac{z}{z_0}\right)^n.$$

I tried using binomial theorem to expend the left-hand side of the equality, $$ \left(1-\frac{z}{z_0}\right)^{-j}=\sum_{n=0}^{-j}{-j\choose n}\left(\frac{z}{z_0}\right)^{n} $$ but it does not resemble the paper's final result.

Thank you for your time and help in advance.

Answer 1

We obtain \begin{align*} \color{blue}{\left(1-\frac{z}{z_0}\right)^{-j}}&=\sum_{n=0}^{\infty}\binom{-j}{n}\left(-\frac{z}{z_0}\right)^n\tag{1}\\ &=\sum_{n=0}^{\infty}\binom{n+j-1}{n}\left(\frac{z}{z_0}\right)^n\tag{2}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}\binom{n+j-1}{j-1}\left(\frac{z}{z_0}\right)^n}\tag{3}\\ \end{align*} in accordance with the paper.

Comment:

• In (1) we apply the binomial series expansion.

• In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (3) we use $\binom{p}{q}=\binom{p}{p-q}$.