QUAM Review Question 5

I'd like someone who's done this problem to see if I've made a mistake.

We are asked to find two Laurent series centered at zero for
$$
f(z)=\frac{1}{z(z-4)}.
$$
For the first series, I factor out a $z$, giving
\begin{align}
\frac{1}{z(z-4)}&=\frac{1}{z^2}\frac{1}{(1-4/z)}\\&=\frac{1}{z^2}\sum_{k=0}^\infty\bigg(\frac{4}{z}\bigg)^k\\&=\sum_{k=0}^\infty\frac{4^k}{z^{k+2}}.
\end{align}
This converges for $|4/z|<1\Rightarrow|z|>4$.

For the second series, I factor out a $z-4$, giving
$$
\begin{align}
\frac{1}{z(z-4)}&=\frac{1}{((z-4)+4)(z-4)}\\&=\frac{1}{(z-4)^2}\frac{1}{(1+4/(z-4))}\\&=\frac{1}{(z-4)^2}\frac{1}{(1-(-4/(z-4)))}\\&=\frac{1}{(z-4)^2}\sum_{k=0}^\infty\bigg(\frac{-4}{z-4}\bigg)^k\\&=\sum_{k=0}^\infty\frac{(-1)^k4^k}{(z-4)^{k+2}}.
\end{align}
$$
This converges for $|-4/(z-4)|<1\Rightarrow|z-4|>4$.

I'm unsure if my domains of convergence are correct or even if they are "centered at zero".

You definitely have two Laurent series for $f$. The first one is centered at zero and has the correct specification for its domain of convergence. The second on is centered at $4$, though, rather than at zero.

To get another Laurent series centered at zero, I recommend that you write the function as
$$\frac{1}{z}\cdot\frac{1}{4-z}$$
and focus on finding a power series for the second term in the product.

Pulling out a $4$ initially instead of a $z$ gives
\begin{align}
\frac{1}{z}\cdot\frac{1}{4-z}&=\frac{1}{4z}\cdot\frac{1}{1-z/4}\\&=\frac{1}{4z}\sum_{k=0}^\infty\bigg(\frac{z}{4}\bigg)^k\\&=\sum_{k=0}^\infty\frac{z^{k-1}}{4^{k+1}}.
\end{align}
This converges for $|z/4|<1\Rightarrow|z|<4$.

However, the original function is undefined at $z=0$, so the annulus of convergence is actually $0<|z|<4$.

@dgallimo Looks good. I would point out that your series isn't defined at $z=0$ either, as expected.

I may just be confused, but it seems like we started with $f(z)=\frac{1}{z(z-4)}$, and at some point lost a factor of $-1$ transforming it into $f(z)=\frac{1}{z(4-z)}$. Am I wrong and/or missing something here?

You're absolutely right. There should be a big ol' $-1$ in front of the series.