complex analysis - Finding a Laurent Series involving two poles

Find the Laurent Series on the annulus $1 < |z| < 4$ for

$$R(z) = \frac{z+2}{(z^2-5z+4)}$$

So I am having a few issues with this. I know there are two poles in this problem particulaly $z = 1$ and $z = 4$, so if I factor out I get it into a form as:

$$ \frac{z+2}{(z-1)(z-4)} $$

and here is where it get's a little hazy. I know there is a relationship in which I would have to split this expression into partial fractions:

$$ \frac{z+2}{(z^2-5z+4)} = \frac{2}{z-4} + \frac{-1}{z-1} $$

now the textbook goes on about using their geometric series, which I somewhat see, but I cannot understand how to get the coefficients. I was trying to use the method of treating each of the numerators in the partial fractions as power series and then solving for coeffecients, but that resulted to no avail. Then I tried not even expanding the expression into partial fractions and attempting to solve for the coefficients by accounting for each singularity and using the remaining part as a power series i.e

$ \frac{z+2}{(z-4)} $ as one power series and then $ \frac{z+2}{(z-1)} $ as the other.

Still not working out. Perhaps my ideas are scattered.

Answer

You need put each partial fraction into the form $\frac{1}{1-w}$ where $|w| \lt 1$ in order to use the geometric series expansion.

$\frac{2}{z-4}$ is analytic in $|z| \lt 4$ and $\Big|\frac{z}{4}\Big| \lt 1$ so we have:

$$
\frac{2}{z-4} = -\frac{1}{2}\cdot\frac{1}{1-\frac{z}{4}} = -\frac{1}{2}\sum_{n=0}^{\infty}\frac{z^n}{4^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{2n+1}}
$$

$-\frac{1}{z-1}$ is analytic in $|z| \gt 1$ and $\Big|\frac{1}{z}\Big| \lt 1$ so we have:
$$
-\frac{1}{z-1} = -\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}\sum_{n=0}^{\infty}\frac{1}{z^n} = \sum_{n=0}^{\infty}-\frac{1}{z^{n+1}}
$$

So in total we have:

$$
R(z) = \dots -\frac{1}{z^3}-\frac{1}{z^2}-\frac{1}{z} -\frac{1}{2}-\frac{z}{8}-\frac{z^2}{32}-\dots
$$