# Laurent Series in powers of $z$ and $\frac{1}{z}$

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
$$a_k = \sum_{|z_j for $$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?

• 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 $-1/(2(z-1))+3/(2(z-3))$. – Lubin Dec 10 '18 at 1:57
• 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. – 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.