Laurent Series for different domains

Question

I would like to determine the Laurent series for the function

$$f(z) = \frac{1}{z(z-2)^3}$$

for two different domains

$$\vert{z}\vert<2$$

and

$$\vert{z}\vert>2$$

But, I am unsure when I am computing the Laurent series when I am to take into consideration the different domains. I use standard tricks such as taking out the $\frac{1}{z}$ term and rearranging the remaining term into the sum formula of the geometric series to compute the general Laurent series.

Thanks for all help!

Answer 1

For $|z|>2$ :

$$f(z) = \frac{1}{z(z-2)^3} = \frac{1}{z[z^3(1-\frac{2}{z})^3]}=\frac{1}{z^4(1-\frac{2}{z})^3}$$

A known geometric series is :

$$\frac{1}{1-w} = \sum_{n=0}^\infty w^n ,\quad |w| < 1$$

From that, we can derive :

$$\frac{1}{(1-w)^3} = \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)w^n, \quad |w|<1$$

So, applying that for $w=\frac{2}{z}$, we get :

$$f(z) = \frac{1}{z^4} \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)\bigg(\frac{2}{z}\bigg)^n=\sum_{n=0}^\infty2^{n-1}(1+n)(2+n)z^{-(n+4)}$$

for $|\frac{2}{z}| < 1 \Leftrightarrow |z| > 2 . $

For $|z|<2$, we :

$$f(z)=\frac{1}{z(z-2)^3}= - \frac{1}{z(2-z)^3}= - \frac{1}{z}\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^n$$

$$=$$

$$-\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^{n-1}$$

which hols for $|z|<2$.