# Expanding fractions as powers of $z$

I'm reading through complex functions in Boas' book, and there's a part when discussing Laurent series where she says:

"Now, for $0 <|z|<1$, we expand each of the fractions in the parenthesis in powers of $z$."

The equation she refers to is the following:

$$f(z) = \frac {4}{z} \left({\frac{1}{1+z}}+ {\frac{1}{2-z}}\right).$$

As a result of the expansion, she gets:

$$f(z)=-3+9z/2-15z^2/4+33z^3/8+ \cdots +6/z.$$

I have no clue how she got the second equation from the first. Specifically, I don't know what she means by "expand each of the fractions in the parenthesis in powers of $z$". An explanation would be appreciated.

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Are you familiar with the geometric series formula? It is $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n;$$ (you need to change the sign) for the second fraction, you can use $$\frac{1}{2-z}=\frac{1}{2}\frac{1}{1-z/2}$$ and then expand. Put the two series together and voilà. – anon Jul 18 '12 at 3:31
Gotit. Thanks to both posters. – Joebevo Jul 18 '12 at 3:54

$$\frac{1}{a \pm z} = \frac{1}{a} \frac{1}{1 \pm \dfrac{z}{a}} = \frac{1}{a}\sum_{n = 0}^{\infty} \left( \frac{\pm z}{a} \right )^n$$