
laurent series
Can someone show me how to find the Laurent series of 1/z(z4) expanded in terms of z1? I can figure out the expansion for the domain 1 < abs(z) < 3, but I'm having trouble with the other two domains (i.e. 0 < abs(z) < 1 and abs(z) > 3.
Thanks for your help.

The has singularities at z=0 and z=4 so you will be able to find two series. One for 0<z<4 and for z>4. Where did you get your values?

It's expanded in terms of z1, no?

The objective is to get the expressions in terms of a geometric series of the form $\displaystyle \frac{1}{1f(z)}$ where $\displaystyle f(z)<1$ in the domain of interest. First write the function as $\displaystyle f(z)=\frac{1}{4z}+\frac{1}{4(z4)}$. For the domain $\displaystyle z1<1$ start by adding and subtracting one in the denominator:
$\displaystyle f(z)=\frac{1}{4}\left(\frac{1}{1+z1}\right)+\frac{1}{4}\left(\frac{1}{3+z1}\right)$
$\displaystyle =\frac{1}{4}\left(\frac{1}{1((z1))}\right)\frac{1}{12}\left(\frac{1}{1\frac{z1}{3}}\right)$
$\displaystyle =1/4\sum_{n=0}^{\infty}(1)^n(z1)^n1/12\sum_{n=0}^{\infty}\frac{(z1)^n}{3^n};\quad z1<1$
Now in the case of $\displaystyle z1>3$, you need to have geometric terms which are less than one for $\displaystyle z1>3$. I'll do the first term:
$\displaystyle \frac{1}{4}\frac{1}{z}=\frac{1}{4}\left(\frac{1}{z1+1}\right)=\frac{1}{4}\left(\frac{1}{(z1)(1+\frac{1}{z1})}\right)$
$\displaystyle =\frac{1}{4}\sum_{n=0}^{\infty}\frac{(1)^n}{(z1)^{n+1}}$
Do the second one to get: $\displaystyle \frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n}{(z1)^{n+1}}$
Thus:
$\displaystyle f(z)=\frac{1}{4}\sum_{n=0}^{\infty}\frac{(1)^n}{(z1)^{n+1}}+\frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n} {(z1)^{n+1}},\quad z1>3$

The question he asks for z>3 not z1>3. That is all i was saying.