finding Laurent series for $\dfrac{1}{z(z-2)^3}$

Question

I am trying to get the Laurent series for $\dfrac{1}{z(z-2)^3}$. I know there are poles at $z = 0$ and $z=2$, and so I am looking for expansions about the singularities. Using $\dfrac{1}{1-z} = \sum_{n=0}^\infty z^n$ for $|z|<1$ and $\dfrac{1}{1-z} = \sum_{n=0}^\infty \dfrac{1}{z^n}$ for $|z|>1$, I find that about $z=0$, $f(z) = \dfrac{-1}{8z} \bigg( 1 + \dfrac{z}{2} + \dfrac{z^2}{4} + ... \bigg)^3$ and about $z = 1$, $f(z) = \dfrac{-1}{8z} \bigg( \dfrac{2}{z} + \dfrac{4}{z^2} + ... \bigg)^3$. I end up cubing the series because I am using the above expansion and trying to get around expanding for the full term. How can I do the Laurent expansions so as not to end up cubing a simpler expansion?

Answer 1

Instead of using the geometric series directly and cubic it, use the fact that $$ \frac{1}{(1-z)^3} = \frac{1}{2}\frac{d^2}{dz^2}\Big(\frac{1}{1-z}\Big) $$ and differntiate the series term by term.

Answer 2

Use the generalised binomial theorem to get $f(z)=\frac{-1}{8z}(1-\frac{z}{2})^{-3}=\frac{-1}{8z}\sum\limits_{i=0}^{\infty}{\binom{-3}{i}(-\frac{z}{2})^i}=\sum\limits_{i=0}^{\infty}{(-\frac{(i+1)(i+2)}{2^{i+4}}z^{i-1})}$ with convergence when $|z|<2$ and $z \not =0$.

and $f(z)=\frac{1}{2(z-2)^3}(1+\frac{z-2}{2})^{-1}=\frac{1}{2(z-2)^3}\sum\limits_{i=0}^{\infty}{(-\frac{z-2}{2})^i}=\sum\limits_{i=0}^{\infty}{\frac{(-1)^i}{2^{i+1}}(z-2)^{i-3}}$ with convergence when $|z-2|<2$ and $z \not =2$.