Finding the Laurent representation of a complex function

Question

How can i find the Laurent representation fot the function: $$f(z)=\dfrac{1}{1-z^2}+\dfrac{1}{3-z}$$

In the region of:

a) $\{z\in\mathbb C:1<|z|<3\}$

b) $\{z\in\mathbb C:1<|z-2|<3\}$

Answer 1

$\frac{1}{1 - z} = \sum z^n$--the simplest function. You have the following:

$$ \frac{1}{2}\frac{1}{1 - (-z)} + \frac{1}{2}\frac{1}{1 - z} + \frac{1}{3}\frac{1}{1 - \frac{z}{3}} $$

When $1 < |z| < 3$, we need the Laurent series for $\frac{1}{1 \pm z}$ but the Taylor series for $\frac{1}{3 - z}$:

$$ \frac{1}{1 - z} = \frac{1}{z}\frac{1}{z^{-1} - 1} = -z^{-1}\frac{1}{1 - z^{-1}} = -z^{-1}\sum_0^\infty z^{-n} = -\sum_1^\infty z^{-n} $$

Which gives:

\begin{align} \frac{1}{1 - z} =&\ -\sum_1^\infty z^{-n} \\ \frac{1}{1 - (-z)} =&\ -\sum_1^\infty (-1)^n * z^{-n} \\ \frac{1}{1 - \frac{z}{3}} =&\ \sum_0^\infty \left(\frac{z}{3}\right)^n \end{align}

Adding this gives something to the affect of:

$$ a_n = \begin{cases} \frac{1}{3} & n = 0 \\ -z^{-n} + \frac{1}{3}\frac{z^n}{3^n} & n \text{ even} \\ \frac{1}{3}\frac{z^n}{3^n} & n \text{ odd} \end{cases} $$

I'm not quite clear on the domain of the second part.

p.s. You actually don't need the partial fraction decomposition (and I believe there was a previous answer which did not use partial fractions).

We know that $\frac{1}{1 - z^2}$'s Taylor series has a radius of convergencce of $|z| < 1$ (and is clearly undefined for $z = \pm 1$), and finally since the Laurent series for $\frac{1}{1 - z} = -\sum_1^\infty z^{-n}$, meaning:

$$ \frac{1}{1 - z^2} = -\sum_1^\infty z^{-2n} \text{, for } |z| > 1 $$

Adding to the Taylor series we get:

$$ -\sum_1^\infty z^{-2n} + \frac{1}{3}\sum_0^\infty \left(\frac{z}{3}\right)^n $$

...but the exponents don't match up (which would be nice)...so match them up by splitting the Taylor series into two sums:

$$ -\sum_1^\infty z^{-2n} + \frac{1}{3}\left(\sum_0^\infty \left(\frac{z}{3}\right)^{2n} + \sum_0^\infty \left(\frac{z}{3}\right)^{2n + 1} \right) $$

Which can be written as:

$$ \frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_0^\infty\left(\frac{z}{3}\right)^{2n + 1}\\ \text{or} \\ \frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_1^\infty\left(\frac{z}{3}\right)^{2n - 1} \\ \text{or}\\ \frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n} + \frac{1}{3}\left(\frac{z}{3}\right)^{2n - 1}\right) $$