Laurent series of $\frac{1}{(z^2 + 1)(z^2 - 9)}$

Question

I want to derive Laurent series of $$f(z) = \frac{1}{(z^2 + 1)(z^2 - 9)}$$

for two sets: $1 < |z| < 3$ and $3 < |z|$.

My work so far

First thing to do is to apply partial fraction decomposition:

$$f(z) = \frac{\frac{-1}{10}}{z^2 + 1} + \frac{\frac{1}{60}}{z - 3} + \frac{\frac{-1}{60}}{z + 3}$$

Now for $1 < |z| < 3$:

$$\frac{1}{z - 3} = -\frac{1}{3 - z} = \frac 1 3 \cdot \frac{-1}{1 - \frac{z}{3}} = - \frac 1 3 \sum_{n = 0}^\infty (\frac z 3)^n$$

$$\frac{1}{z + 3} = \frac 1 3 \cdot \frac{1}{1 + \frac z 3} = \frac 1 3 \cdot \frac{1}{1 - (- \frac z 3)} = \frac 1 3 \sum_{n = 0}^\infty (-1)^n (\frac z 3)^n$$

So $$f(z) = \frac{-\frac{1}{10}}{z^2 + 1} + \frac{1}{60} (-\frac 1 3 \sum_{ n = 0}^\infty (\frac z 3)^n) -\frac{1}{60} \cdot \frac 1 3 \sum_{n = 0}^\infty (-1)^n (\frac z 3)^n$$

Now for $|z| > 3$

$$\frac{1}{z - 3} = \frac 1 z \cdot \frac{1}{1 - \frac 3 z} = \frac 1 z \sum_{n = 0}^\infty (\frac 3 z)^n$$

$$\frac{1}{z + 3} = \frac 1 z \cdot \frac{1}{1 - (- \frac 3 z)} = \frac 1 z \sum_{n = 0}^\infty (-1)^n \cdot (\frac 3 z)^n$$

So $$f(z) = \frac{-\frac{1}{10}}{z^2 + 1} + \frac 1 z \sum_{n = 0}^\infty (\frac 3 z)^n -\frac 1 z \cdot \frac{1}{1 - (- \frac 3 z)} - \frac{1}{60} \cdot \frac 1 z \sum_{n = 0}^\infty (-1)^n \cdot (\frac 3 z)^n$$$

I didn't expand $\frac{1}{z^2 + 1}$ because in my opinion it cannot be expand since it has no real roots. Am I correct with my calculations?

Answer 1

I am myself still in the course complex analysis, so feel free to correct me if I'm wrong as I'm not sure of my answer.

For Laurent series, it is useful to remember the following two geometric series, \begin{align} \frac{1}{1-cz} &= \sum_{n=0}^\infty (cz)^n, \text{ for } |z|<\frac{1}{|c|},\\ \frac{1}{1-\frac{b}{z}} &= \sum_{n=0}^\infty (\frac{b}{z})^n, \text{ for } |z|>|b|. \end{align}

Note that $\frac{1}{z^2-1} = \frac{1}{z-i} + \frac{1}{z+i}$. So we can find the Laurent series for each of these.

\begin{align} \frac{1}{z-i} &= \frac{1}{z}\frac{1}{1-\frac{i}{z}}\\ &= \frac{1}{z}\sum_{n=0}^\infty (\frac{i}{z})^n, \text{ for } |z|>1=|i|\\ &= \sum_{n=0}^\infty \frac{i^n}{z^{n+1}}, \text{ for } |z|. \end{align}

Then the others can be calculated in a similar way, using the geometric series.