As title suggests, I want to find a Laurent series expansion for $$\frac{1}{z^2-4z+3}=\frac{1}{(z-3)(z-1)}=\frac{1}{2}\frac{1}{z-3}-\frac{1}{2}\frac{1}{z-1}$$ about $z=0$ with $|z|<1$. We have singularities at $z=1$, and $z=3$. I am confused about whether I want a Taylor series for both terms. That is, is the series expansion given by the following:
$$\frac{1}{2}\sum_{0}^{\infty}\frac{z^n}{3^n}-\frac{1}{2}\sum_{0}^{\infty}z^n?$$ I'm somewhat confused about the region. Both singularities lie outside the center, so they should both be Taylor series expansions, but I'm not so sure.
Best Answer
\begin{align*} \dfrac{1}{2}\dfrac{1}{z-3}&=-\dfrac{1}{6}\dfrac{1}{1-(z/3)}\\ &=-\dfrac{1}{6}\sum_{n=0}^{\infty}\left(\dfrac{z}{3}\right)^{n}, \end{align*} note that $|z|<1$ implies that $|z/3|<1/3<1$. \begin{align*} -\dfrac{1}{2}\dfrac{1}{z-1}&=\dfrac{1}{2}\dfrac{1}{1-z}\\ &=\dfrac{1}{2}\sum_{n=0}^{\infty}z^{n}, \end{align*} so the whole sum is \begin{align*} \dfrac{1}{6}\sum_{n=0}^{\infty}\left[3z^{n}-\left(\dfrac{z}{3}\right)^{n}\right]&=\dfrac{1}{6}\sum_{n=0}^{\infty}\left(3-\dfrac{1}{3^{n}}\right)z^{n}. \end{align*}