0
$\begingroup$

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 $r <|z| < R$

Where
$a_k = \sum_{|z_j<r|} z_j^{k-1} Res(f; z_j)$ 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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

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.