# Power series by partial fractions

I am trying to find a power series centered at the origin for the function $f(z) = \frac {1}{1-z-2z^2}$ by first using partial fractions to express $f(z)$ as a sum of two simple rational functions. If I am not mistaken, I have the following: $\frac {A}{-2z+1} + \frac {B}{z+1} \Rightarrow A(z+1) + B(-2z+1) = 1$. From here, I deduced that $A= \frac{2}{3}$, and $B=\frac{1}{3}.$ If that is correct, then here is where I get stuck: I know that if $f$ is analytic in $D(\alpha; r)$, then there exists constants $C_k$ such that $f(z)=\sum_{k=0}^{\infty} C_k (z-\alpha)^k$ for all $z \in D(\alpha;r)$. Hence, I would write $f(z) = \frac{\frac{1}{3}}{-2z+1} + \frac{\frac{2}{3}}{z+1} \Rightarrow \frac{\frac{1}{3}}{-2z+1} + \frac{\frac{2}{3}}{z+1} = \sum_{k=0}^{\infty} C_k (z-\alpha)^k$. Am I missing a step or have I not simplified the partial fraction decomposition appropriately? One such simplification I could think of is multiplying by the reciprocal of the denominators to give me $\frac{1}{-6z+3} + \frac{2}{3z+3}$. However, then I am still confused about how to proceed from here. Any help is greatly appreciated.

Source: Complex Analysis, Third Edition by Joseph Bak and Donald J. Newman.

I think you are overthinking this. Your partial fractions are correct. Now just use a geometric series on each:

$$\frac{1}{1-2z} = \sum_{k=0}^{\infty} (2 z)^k$$

$$\frac{1}{1+z} = \sum_{k=0}^{\infty} (-z)^k$$

Now put these together into a single sum. You should get something like

$$\frac{1}{1-z-2 z^2} = \sum_{k=0}^{\infty} a_k z^k$$

$$a_k = \frac{2}{3} 2^k + \frac{1}{3} (-1)^k$$

• Thank you very much for the information. However, I just want to ask how you get $(2z)^k$ and $(-z)^k$ on the two sums above. I think I have a much clearer idea of what is going on, but would like an explanation of how those two. From what I recall about geometric series, for $-1 < r < 1$, the geometric series satisfies $\sum_{n=0}^{\infty} r^n = \frac {1}{1-r}$. With this in mind, is $r$ replaced with the complex variable $z$ from the original function? – Jamil_V Mar 29 '13 at 21:03
• @user67418: I think you just said it yourself by writing down the generic geometric series. Yes, I just replaced your $r$ with $2 z$ and $-z$, respectively. – Ron Gordon Mar 29 '13 at 21:19

Each term is representable by a power series by using the geometric series theorem. Notice that $${1\over 1 - 2x} = \sum_{n=0}^\infty 2^n x^n, \qquad |x| < 1/2,$$ and that $${1\over 1 + x} = \sum_{n=0}^\infty (-1)^n x^n, \qquad |x| < 1.$$ Combine appropriately to get your desired power series expansion.

• Thank you for your response; I greatly appreciate it. – Jamil_V Mar 29 '13 at 21:40