0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

\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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.