How to use Laurent expansion of a function into this form?

Question

How can I compute the Laurent expansion of the function $$f(z)=\frac{1}{(2z+1)(z-2)}$$ when $\frac{1}{2}<|z|<2$ with this $\frac{1}{z^2}$ term?

Answer 1

So this is very simple;

first you compute the partial fraction decomposition:

$$\frac{1}{\left(2x+1\right)\left(x-2\right)} = \frac{A}{2x+1}+\frac{B}{x-2}$$

which gives you A = -2/5 and B=1/5

so:

$$\frac{1}{\left(2x+1\right)\left(x-2\right)}=-\frac{2}{5}\cdot\frac{1}{2x+1}+\frac{1}{5}\cdot\frac{1}{x-2}$$

and then you put those into the geometric series:

$$-\frac{2}{5}\cdot\frac{1}{2x+1}+\frac{1}{5}\cdot\frac{1}{x-2}=-\frac{2}{5}\sum_{n=0}^{\infty}\left(-2x\right)^{n}+\frac{1}{5}\cdot\left(-\frac{1}{2}\right)\sum_{n=0}^{\infty}\left(\frac{1}{2}x\right)^{n}=\sum_{n=0}^{\infty}\left(-\frac{2}{5}\cdot\left(-2\right)^{n}-\frac{1}{10}\cdot\left(\frac{1}{2}\right)^{n}\right)x^{n}$$

Answer 2

Maybe this is the solution: $$\begin{align}\frac1{\left(2z+1\right)\left(z-2\right)} &=-\frac25\cdot\frac1{2z+1}+\frac15\cdot\frac1{z-2} \\&=-\frac1{5z}\cdot\frac1{1+\frac1{2z}}-\frac1{10}\cdot\frac1{1-\frac z2}\\ &=-\frac1{5z}\sum_{n=0}^\infty\left(\frac{-1}{2z}\right)^n-\frac1{10}\sum_{j=0}^{\infty}\left(\frac z2\right)^j\\&=\sum_{j=1}^\infty\left(\frac{(-1)^j}{5\cdot2^{j-1}}\frac1{z^j}\right)-\sum_{j=0}^\infty\left(\frac1{10\cdot2^j}z^j\right)\end{align}$$ Visual Complex Analysis-Annular Laurent Series