Finding Laurent series where given annulus is not in a singularity

Question

I'm given a problem where I need to calculate the Laurent series of $f(z)$ inside the given annulus

$$ f(z) = {1\over z^3(z-1)}; \quad 1 < |z| < 2 $$

From online resources(videos, notes) I understand that I need a product of:

Laurent series in $z_0=0$ valid for $1<|z|$.
Taylor series in $z_0=0$ valid for $|z|<2$.

I caculated the Laurent series from $f(z)$ and got

$$ \sum_{n=0}^\infty {1\over z^{n+4}}; \quad 1 < |z| $$

However i don't know how to proceed with the Taylor series valid for the region $|z|<2$. (I got a Taylor series but is valid for the region $|z| < 1$).

Answer 1

You're already done: if a series is valid on $1 < |z|$, then it's already valid inside $1 < |z| < 2$, because everything in that annulus satisfies $1 < |z|$.

Because there are no singularites with $|z| = 2$, there does not exist a Laurent series for $f(z)$ that converges only in $1 < |z| < 2$. Due to the placement of the singularities of $f(z)$, the only possible domains of convergence of a Laurent series for $f$ are:

• $0 < |z| < 1$, possibly along with some boundary points
• $1 < |z| < \infty$, possibly along with some boundary points

I don't know the method you're trying to describe from your online resources, but I can guess at it from your description. These sources suggest factoring your function into two functions

$$ f(z) = f_1(z) f_2(z) $$

with the property that $f_1$ has no singularities in $1 < |z| < \infty$ and $f_2$ has no singularities in $|z| < 2$. For example, this could be done by choosing any $f_1$ that has no singularities on $1 < |z| < \infty$ and has the same poles that $f(z)$ does on $|z| \leq 1$, and then simply setting $f_2(z) = f(z) / f_1(z)$.

Then, you can find a Laurent series for $f_1(z)$ valid on $1 < |z|$ and a Taylor series for $f_2(z)$ valid on $|z| < 2$, and if you multiply them, you would get a Laurent series for $f(z)$ valid on $1 < |z| < 2$.

For this problem, the easiest factorization is $f_1(z) = f(z)$ and $f_2(z) = 1$.