$f(z)$ is defined like this:
$$
f(z) = \frac{z}{(z-1)(z-3)}
$$
I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 \leq |z – 1| \leq 2$.
What I understand from question is I must expand $f(z)$ Laurent series.
$$ f(z) = \sum_{m=0}^{\infty}a_{m}(z-1)^{m} + \sum_{m=1}^{\infty}b_{m}(z-1)^{-m}$$
where,
$$ a_{m} = \frac{1}{j2\pi}\oint_{C}\frac{f(z)}{(z-1)^{m+1}}dz $$
$$ b_{m} = \frac{1}{j2\pi}\oint_{C}\frac{f(z)}{(z-1)^{1-m}}dz $$
This is what theory tells me.
But I apply partial fraction method to this function like this:
$$ f(z) = \frac{z}{(z-1)(z-3)} = \frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = \frac{-1/2}{(1-z^{-1})} + \frac{1/2}{(1-3z^{-1})} $$
And I know this series expansion from z-transform like this:
$$ f(z) = -\frac{1}{2} \sum_{k=0}^{\infty}z^{-k} + \frac{1}{2} \sum_{k=0}^{\infty}3^{k}z^{-k}$$
I obtain a series expansion but it looks like Mclaurin series not a Laurent series.
Here, my first question an expression may have different type of series expansion?
And second, how to find a Laurent series for $ f(z) $
Best Answer
The problem is that if you use $\frac 1{1-z} = \sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!
Hence the $\frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $\frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).
You have to expand in a neighborhood of $1$ the expression $\frac z{z-3}$. You can set $t = z - 1 \implies z = t + 1$ so your expression becomes $\frac{t+1}{t-2}$.
Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $\frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.