Laurent series expansion of $\frac{1}{z(z-2)^2}$ at $z=2$

Question

Im struggling a bit with Laurent series expansions, and I am stuck on the following exercise:

Find a Laurent series expansion with center $z=2$ of $f(z)=\frac{1}{z(z-2)^2}$

My solution so far is not a lot:

I've found that there will be one Laurent series at the punctured disk $D(2,2)$ and one $A=(2,2,\infty)$

Starting with the punctured disk $D(2,2)$ I know I want the polynomial to have the form $\frac{1}{1-c*z}$ since we are expanding in our "inner" area.

The part im stuck on is how I should get $\frac{1}{z(z-2)^2}$ = $\frac{1}{z}\frac{1} {(z-2)^2}$ to $\frac{1}{1-c*z}$ to able to write as a series.

Can anyone help me out please?

Thanks!

Answer 1

Using partial fraction expansion, we can write

$$\begin{align} \frac1{z(z-2)^2}=\frac1{4z}-\frac1{4(z-2)}+\frac1{2(z-2)^2} \end{align}$$

---

---

To obtain the Laurent series for $|z-2|<2$, we write

$$\begin{align} \frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\ &=\frac18 \frac1{1+\frac{z-2}2}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\ &=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac18\sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n \end{align}$$

---

---

And to obtain the Laurent series for $|z-2|>2$ we write

$$\begin{align} \frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\ &=\frac1{4(z-2)} \frac1{1+\frac2{z-2}}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\ &=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac1{8}\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n\\\\ &=\frac1{8}\sum_{n=3}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n \end{align}$$

Answer 2

This becomes simpler to think about if you make a substitution $w=z-2$, so you are just looking to expand $\frac{1}{w^2(w+2)}$ in terms of $w$. Now you can just write it as $\frac{1}{w^2}\cdot\frac{1}{w+2}$ where the second factor can be expanded as a geometric series $\frac{1}{2}\frac{1}{1+w/2}=\frac{1}{2}\sum_{n=0}^\infty(-w/2)^n$ near $w=0$ (i.e., near $z=2$).

(For the Laurent expansion near $\infty$, you can use a similar approach but instead substitute $w=\frac{1}{z-2}$.)