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Find the Laurent expansion for the function $f(z)$ in the domains $i),ii)$

$f(z)=\frac{1}{(z-1)(z-2)(z-3)}$

$i)$ $0<|z|<1$

$ii)$ $1<|z|<2$

What I did for the first:

$f(z)=\frac{1}{(z-1)(z-2)(z-3)}=\frac{-1}{(z-2)}+\frac{1}{2(z-3)}-\frac{1}{2}\sum_{n=0}^\infty z^n$

Since $|z|<1$

Now, I don't know if this is correct, and also, I don't know how to do $ii)$

  • 1
    [WolframAlpha](http://www.wolframalpha.com/input/?i=partial+fraction+decomposition+for+1%2F((z-1)(z-2)(z-3))) agrees with your PFD with the exception of the term $\frac{1}{(z-3)}$, which wolfram says is $\frac{1}{2(z-3)}$. – 2017-01-20

2 Answers 2

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$f(z)=\frac{1}{(z-1)(z-2)(z-3)}=\frac{-1}{(z-2)}+\frac{1}{2(z-3)}-\frac{1}{2(1-z)}$

If $1<|z|<2$,

$f(z)=\frac{1}{2(1-z/2)}-\frac{1}{6(1-z/3)}-\frac{1}{2z(1-1/z)}$

Clearly, each of the three geometric series converge if $|z/2|<1$, $|z/3|<1$ and $|1/z|<1$ which is supported by the fact $1<|z|<2$.

$f(z)=(1/2)\sum_0^\infty (z/2)^n-(1/6)\sum_0^\infty (z/3)^n-(1/2)\sum_0^\infty (1/z)^{n+1}$

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1. Laurent expasions of $\; f_1(z)=\dfrac{1}{z-1}$

$$\frac{1}{z-1}=\begin{cases} -\dfrac{1}{1-z}=-\displaystyle\sum_{n=0}^{+\infty}z^n,\quad |z|<1,\quad (L_1) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{1}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{1}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{1}{z^{n+1}},\quad |z|>1,\quad (L_2)\end{cases}$$ 2. Laurent expasions of $\;f_2(z)=\dfrac{1}{z-2}$ $$\frac{1}{z-2}=\begin{cases} -\dfrac{1}{2}\dfrac{1}{1-\dfrac{z}{2}}=-\dfrac{1}{2}\displaystyle\sum_{n=0}^{+\infty}\left(\dfrac{z}{2}\right)^n=-\displaystyle\sum_{n=0}^{+\infty}\dfrac{z^n}{2^{n+1}},\quad |z|<2,\quad (L_3) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{2}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{2}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{2^n}{z^{n+1}},\quad |z|>2,\quad (L_4)\end{cases}$$

3. Laurent expasions of $\;f_3(z)=\dfrac{1}{z-3}$

$$\frac{1}{z-3}=\begin{cases} -\dfrac{1}{3}\dfrac{1}{1-\dfrac{z}{3}}=-\dfrac{1}{3}\displaystyle\sum_{n=0}^{+\infty}\left(\dfrac{z}{3}\right)^n=-\displaystyle\sum_{n=0}^{+\infty}\dfrac{z^n}{3^{n+1}},\quad |z|<3,\quad (L_5) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{3}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{3}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{3^n}{z^{n+1}},\quad |z|>3,\quad (L_6)\end{cases}$$

4. Laurent expasions of $\;f(z)=\dfrac{1}{(z-1)(z-2)(z-3)}=\dfrac{1}{2}f_1(z)-\dfrac{1}{2}f_2(z)+\dfrac{1}{2}f_3(z).$

(i) $\;f(z)=\dfrac{1}{2}(L_1)-\dfrac{1}{2}(L_3)+\dfrac{1}{2}(L_5)\quad \text{ valid for }0<|z|<1$

(ii) $\;f(z)=\dfrac{1}{2}(L_2)-\dfrac{1}{2}(L_3)+\dfrac{1}{2}(L_5)\quad \text{ valid for }1<|z|<2$

Etc. for other possible regions.