I am very confused! I am trying to Expand a given function in a Laurent series valid for different values of z for a complex variables class. I tried going to my professor for help and she was able to help for part b and gave me some good hints for c and d, however I am still struggling to figure this out.
The questions states:
expand $f(z) = \frac{z}{(z-1)(2-z)}$ in a Laurent series valid for:
- a) $\quad|z|<1$
- b) $\quad1<|z|<2$
- c) $\quad|z|>2$
- d) $\quad|z-1|>1$
- e) $\quad0<|z-2|<1$
Find the Laurent expansion of f (z) = 1/( z(z − 1)(z − 2)), (in powers of z) for: a. 0 < |z| < 1 b. 1 < |z| < 2 c. |z| > 2.
I found this example helpful because it attempts to solve for part b of this user's question (I think) using the same process that I am trying, however I'm confused why you just consider one part of the partial fraction as opposed to both.
For part b) I was told to do it for both parts of the partial fraction and then add the resulting Laurent series together.
What I came up with was the sum of $\left(\frac{z}{2}\right)^n$ for $n=[0,\infty]$ + the sum of $\frac{1}{z^n}$ for $n=[1,\infty]$.
So when I try and do this for part A and C (i.e. only considering one half of the partial fraction) I get confused because I'm getting an incorrect answer.
If someone could work one of these through and explain what they are doing I would really appreciate it!
The answers provided by the back of the book are:
- a) $$-\frac{z}{2}-\frac{3z^2}{4}-\frac{7z^3}{8}-\frac{15z^4}{16}$$
- b) $$\cdots+\frac{1}{z^2}+\frac{1}{z}+1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots$$
- c) $$-\frac{1}{2}-\frac{3}{z^2}-\frac{7}{z^3}-\frac{15}{z^4}-\cdots$$
(this one is especially confusing to me because it seems like its missing the first order z term)
- d) $$-(z-1)^{-1} – 2(z-1)^{-2} – 2(z-1)^{-3} – \ldots$$
- e) $$1-2(z-2)^{-1}-(z-2)^{-1}-(z-2)+(z-2)^2-(z-2)^3+(z-2)^4-\ldots$$
Best Answer
The idea is basically what Cameron already mentioned in his reply. First, use partial fractions to expand $f(z)$
$$ f(z) = \frac{z}{(z-1)(2 - z)} = z\bigg(\frac{a}{z - 1} + \frac{b}{2 - z}\bigg) = z\bigg(\frac{z(b - a) + (2a - b)}{(z-1)(2 - z)}\bigg) , $$
which implies $a=b=1$,
$$ f(z) = \frac{z}{z - 1} + \frac{z}{2 - z} $$
Before we go to your questions remember that if $|x|<1$ then
$$ \frac{1}{1 - x} = \sum_{k = 0}^{\infty} x^k $$
$|z| < 1$
In this region $|z/2| = |z|/2 < 1/2 < 1$, therefore
\begin{eqnarray*} f(z) &=& -z\frac{1}{1 - z} + \frac{z}{2}\frac{1}{1 - z/2}\\ &=& -z\sum_{k = 0}^{\infty} z^k + \frac{z}{2} \sum_{k = 0}^{\infty} \left(\frac{z}{2}\right)^k = \sum_{k = 0}^{\infty}\left( \frac{1}{2^{k + 1}} - 1\right)z^{k+1} \end{eqnarray*}
$1< |z| < 2$
In this case note that $|1/z| < 1$, and $|z/2| < 1$, then
\begin{eqnarray*} f(z) &=& \frac{1}{1 - 1/z} + \frac{z}{2}\frac{1}{1 - z/2}\\ &=& \sum_{k = 0}^{\infty} \frac{1}{z^k} + \frac{z}{2} \sum_{k = 0}^{\infty} \left(\frac{z}{2}\right)^k = \sum_{k = 0}^{\infty}\bigg(\frac{1}{z^{k}} + \frac{1}{2^{k + 1}}z^{k+1} \bigg) \end{eqnarray*}
$|z| > 2$
For this region $|1/z| < 1/2 < 1$ and $|2/z| < 1$ so that
\begin{eqnarray*} f(z) &=& \frac{1}{1 - 1/z} - \frac{1}{1 - 2/z}\\ &=& \sum_{k = 0}^{\infty} \frac{1}{z^k} - \sum_{k = 0}^{\infty} \left(\frac{2}{z}\right)^k = \sum_{k = 0}^{\infty}\left( 1 - 2^{k}\right)\frac{1}{z^k} \end{eqnarray*}
$|z - 1| > 1$
Note that $1 / |z - 1| < 1$
\begin{eqnarray*} f(z) &=& \frac{(z-1)+1}{z - 1} + \frac{(z-1)+1}{1 - (z - 1)} =1+ \frac{1}{z - 1} - \frac{(z-1)+1}{z-1}\frac{1}{1 - 1/(z - 1)}\\ &=&1+ \frac{1}{z - 1} -\frac{1}{1 - 1/(z - 1)} -\frac{1}{z - 1}\frac{1}{1 - 1/(z - 1)}\\ &=& 1+\frac{1}{z - 1} -\sum_{k = 0}^{\infty} \left(\frac{1}{z - 1}\right)^k -\frac{1}{z - 1} \sum_{k = 0}^{\infty} \left(\frac{1}{z - 1}\right)^k = \frac{1}{z - 1} - 2\sum_{k = 1}^{\infty} \frac{1}{(z - 1)^k} \end{eqnarray*}