Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

1 Answer 1

up vote 2 down vote accepted

Basically the author is using the geometric series expansion in each of the two terms inside the parenthesis.

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.