Exericise 8.35: A laurent series and integral

Find a Laurent series for
$$f(z) = \frac{1}{(z^2-4)(z-2)}$$
centered at $z=2$ and specify it's domain of convergence.

Then, use this to compute
$$\int_{C_1(2)}f(z)dz.$$

To find a Laurent series for f(z),it's helpful to rearranged f(z) such that $$f(z)=\frac{1}{(z^2-4)(z-2)}=\frac{1}{(z-2)^2(z+2)}=\frac{1}{(z-2)^2}*\frac{1}{(z-2)+4}$$
$$=\frac{1}{(z-2)^3}*\frac{1}{1+\frac{4}{z-2}}.$$
This gives us our laurent series,
$$=\frac{1}{(z-2)^3}*\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(z-2)^n}=\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(z-2)^{n+3}}.$$
This laurent series therefore converges when $|z-2|>4$ and diverges when $|z-2|<4.$

Is this all there is to finding a Laurent series? In class, we seemed to be doing some fancy sum from $-\infty$ to $+\infty$, and the series itself converged on an annulus. Am I recalling something different?

A Laurent series is any series of the form $\sum_{n \in \mathbb{Z}} a_n z^n$. Including those where many of the $a_n$'s are zero, such as this series. One could write what opernie found as a sum from $-\infty$ to $\infty$, but many of the coefficients would simply be zero.

And, it looks like what he found does converge on an annulus, just one with infinite outer radius.

Furthermore, looking at the series opernie has, it's term of the form
$$
\frac{a}{(z-2)},
$$

which would be what Mark was calling the "minus first" term, has a coefficient of zero. This is the residue of the function. So, using the residue theorem, we can conclude the integral is zero.

Would you not want to factor out as many $\frac{1}{(z-2)}$ terms as needed in order to find your minus first term? $$\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(z-2)^{n+3}}=\frac{1}{(z-2)^2}\sum_{n=0}^{\infty}\frac{(-1)^n4^n}{(z-2)^{n+1}}.$$

No. This is hard because we haven't really covered it yet, but it is only the term in the Laurent series which has the power $(z-z_0)^{-1}$ that contributes to the value of a closed integral around $z_0$. In this case, $z_0=2$, our contour contains $z_0$, and the Laurent series has a coefficient of zero on that term. If we factor things out of the sum, we can create a term of that form, but it's no longer a Laurent series; it's the product of $(z-2)^{-2}$ with some other Laurent series.