I'm working on a few problems from my textbook and have a bit of trouble figuring out a few things.

In a particular case, suppose $R$ is a rational function all of whose poles in the plane have order one and which has no pole at the origin.

After long division:
$R(z) = S(z) + \frac{P(z)}{Q(z)}$

Where $S$, $P$ and $Q$ are polynomials, and the degree of $P$ is strictly less than the degree of $Q$, concentrating on expanding $f = \frac{P}{Q}$

Then we can write:
$f(z) = \sum_{-\infty}^{\infty} a_k z^k$
for $r <|z| < R$

$a_k = \sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ for $k \leq -1$
and $= -\sum_{|z_j|>R} z_j^{-k-1} Res(f; z_j)$ for $k \geq 0$

Now for the question:
Suppose we want to find the Laurent Series expansion for $R(z)$ in the region $|z|<1$
Let $R(z) = \frac{z^3 - 3z^2 + 3} {(z-1)(z-3)}$
After long division we get:
$R(z) = (z + 1) + \frac{z}{(z-1)(z-3)}$

In this case $r = 0$, $R = 1$, so the only terms that are used are from the negative part of $a_k$ so:
$\frac {z} {(z-1)(z-3)} = \sum_{k=0}^{\infty} a_k z^k$, where $a_k = - [-\frac{1}{2} + \frac{3}{2} \frac{1}{3^{k+1}}]$ for $k \geq 0$

I'm confused on how the terms for $a_k$ were found. I know the definition was given, but that's what I'm having trouble with. First off, what does $z_j$ represent? How do we find the Residue before finding the Laurent Series?

  • $\begingroup$ You are presumably looking for a series in powers of $z$. Since your function has no pole at $z=0$, there is no residue, your series is not merely a Laurent series but a regular power series. The specific numbers come from the Partial Fraction expansion of $z/((z-1)(z-3))$ as $-1/(2(z-1))+3/(2(z-3))$. $\endgroup$ – Lubin Dec 10 '18 at 1:57
  • $\begingroup$ After doing the partial fraction decomposition, what would be the next step? I'm confused as I don't fully understand how each of the coefficients are derived. $\endgroup$ – jd94 Dec 10 '18 at 3:25

You asked for a fuller explanation than I gave in my comment. We get the following: \begin{align} \frac z{(z-1)(z-3)}&=\frac z{(1-z)(3-z)}\\ &=\frac{1/2}{1-z}-\frac{3/2}{3-z}=\frac{1/2}{1-z} - \frac{1/2}{1-\frac z3}\\ &=\frac12\Bigl(1+z+z^2+z^3+\cdots\Bigr)\\ &\qquad-\frac12\Bigl(1+\frac z3+\frac{z^2}9+\frac{z^3}{27}+\cdots\Bigr)\,, \end{align} which should agree with your numbers.


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