Laurent series

1. Nov 29, 2016

dykuma

1. The problem statement, all variables and given/known data

and the solution (just to check my work)

2. Relevant equations
None specifically. There seems to be many ways to solve these problems, but the one used in class seemed to be partial fractions and Taylor series.

3. The attempt at a solution
The first step seems to be expanding this using partial fractions, giving me

Now, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z.

This is the Laurent series for f (z) which is valid in the region 0 < |z| < 1. I then need to get the other two series, which the next one I should try to get is for the region |z| > 2. To get that, it is suggested that I write the two partial fractions as:

However I am not sure what to do with this. I have seen things saying I should expand these two functions, and then add them together, however this does not give me the answer for the region |z| > 2, (in fact, it just gives me the first series, but a degree higher, which makes sense).

2. Nov 29, 2016

vela

Staff Emeritus
The reason for writing
$$-\frac{1}{z-1} = -\frac 1z\left(\frac{1}{1-\frac 1z}\right)$$ and then expanding is because when $|z|>1$, you have $|1/z| < 1$ so a series in positive powers of (1/z) will converge. If you look at the way you expanded 1/(z-2), you should see that you'll get a series of positive powers of (z/2), which won't converge for |z|>2.

3. Nov 29, 2016

dykuma

Thanks! Right, I okay, I see what I did wrong there, and I was able to get the correct series solution for |z| > 2.

The only question I have left is how to get the 1< |z| < 2 series. I am not sure where to start there.

4. Dec 7, 2016

arpon

$$\frac{1}{z}\left(\frac{1}{z-2} - \frac{1}{z-1}\right)= \frac{1}{z}\left(\frac{1}{-2}\cdot\frac{1}{1-\frac{z}{2}} - \frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\right)$$
For $1<|z|<2$, we have, $\left|\frac{1}{z}\right|<1$ and $\left|\frac{z}{2}\right|<1$. So you can now use the formula for infinite geometric series.