# laurent series

• Nov 11th 2008, 03:17 PM
PvtBillPilgrim
laurent series
Can someone show me how to find the Laurent series of 1/z(z-4) expanded in terms of z-1? 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.

• Nov 11th 2008, 03:33 PM
whipflip15
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?
• Nov 11th 2008, 04:25 PM
PvtBillPilgrim
It's expanded in terms of z-1, no?
• Nov 11th 2008, 04:42 PM
shawsend
The objective is to get the expressions in terms of a geometric series of the form $\frac{1}{1-f(z)}$ where $|f(z)|<1$ in the domain of interest. First write the function as $f(z)=-\frac{1}{4z}+\frac{1}{4(z-4)}$. For the domain $|z-1|<1$ start by adding and subtracting one in the denominator:

$f(z)=-\frac{1}{4}\left(\frac{1}{1+z-1}\right)+\frac{1}{4}\left(\frac{1}{-3+z-1}\right)$

$=-\frac{1}{4}\left(\frac{1}{1-(-(z-1))}\right)-\frac{1}{12}\left(\frac{1}{1-\frac{z-1}{3}}\right)$

$=-1/4\sum_{n=0}^{\infty}(-1)^n(z-1)^n-1/12\sum_{n=0}^{\infty}\frac{(z-1)^n}{3^n};\quad |z-1|<1$

Now in the case of $|z-1|>3$, you need to have geometric terms which are less than one for $|z-1|>3$. I'll do the first term:

$-\frac{1}{4}\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{z-1+1}\right)=-\frac{1}{4}\left(\frac{1}{(z-1)(1+\frac{1}{z-1})}\right)$

$=-\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{(z-1)^{n+1}}$

Do the second one to get: $\frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n}{(z-1)^{n+1}}$

Thus:

$f(z)=-\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{(z-1)^{n+1}}+\frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n} {(z-1)^{n+1}},\quad |z-1|>3$
• Nov 11th 2008, 05:25 PM
whipflip15
The question he asks for |z|>3 not |z-1|>3. That is all i was saying.