# Thread: Help with a Laurent expansion

1. ## Help with a Laurent expansion

Hello. My book doesn't explain Laurent series expansions very well and so I was hoping for some help figuring out a problem from the exercises. The answers are in the book so I know I got the first two - I just don't understand the last two.

'Find a Laurent series for the function $\displaystyle 1/(z+z^2)$ in each of the following domains:

a. $\displaystyle 0 < |z| < 1$ ...'

For this one I did a partial fraction decomp and got $\displaystyle 1/z - 1/(z+1)$.

Finding a geometric series for the second term ... $\displaystyle 1/(z+1) = 1/(1-(-z))$ and since $\displaystyle |-z| = |z| < 1$, it has a geometric series representation $\displaystyle \displaystyle\sum\limits_{j=0}^\infty (-z)^j$.

So, $\displaystyle 1/(z+z^2) = 1/z - \displaystyle\sum\limits_{j=0}^\infty (-z)^j$ (So because z=0 is not defined in the annulus, 1/z is analytic - so it's OK to use and also has no series representation?) $\displaystyle = \displaystyle\sum\limits_{j=-1}^\infty (-1)^{j+1}z^j$.

That and the domain (b) $\displaystyle 1 < |z|$ I got OK - for (b) I factored out $\displaystyle (1/z)$ from $\displaystyle (1/z+1)$ to get a geometric series.

My problem is when you move the annulus center to the other singularity...

c. $\displaystyle 0 < |z+1| < 1$ { = $\displaystyle - \displaystyle\sum\limits_{j=-1}^\infty (z+1)^j$ } and d. $\displaystyle 1 < |z+1|$ --

For c. I assume you can still use $\displaystyle 1/z$ with no problems since the annulus doesn't includ z=0, but I am having a hard time trying to find a geometric series for $\displaystyle 1/(z+1)$.

Reverse triangle inequality didn't get me $\displaystyle |z| < 1$ (I got $\displaystyle 0 < |z| < 2$ - although I have doubts about that being correct); I don't think I can simply factor out $\displaystyle 1/z$ to get $\displaystyle 1/z*(1/(1+(1/z))$ because it's possible for $\displaystyle |z|<1$ which would make $\displaystyle |1/z|>1$ ... so I just don't know what to try next.

2. If You want to follow a very confortable way, then...

$\displaystyle \frac{1}{z+z^{2}} = \frac{1}{z}\cdot \frac{1}{1+z} = \frac{1}{z} -1 + z - z^{2} + \dots$ (1)

It is easy to verify that (1) converges for $\displaystyle 0 < |z| < 1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. I was actually looking for help in the annulus $\displaystyle 0 < |z+1| < 1$, but I believe I found out how to do it. (Found a pretty good video on YouTube explaining Laurent series--)

$\displaystyle 1/z+z^2 = 1/z * 1/(1+z)$

$\displaystyle 1/(1+z)$ stays as it is since $\displaystyle z=-1$ is the center of the "annulus" (punctured disk). (The 'why' I haven't completely figured out yet.)

For $\displaystyle 1/z = -1/(1-(1+z))$ and we already know $\displaystyle |1+z|<1$, so that gives the geometric series $\displaystyle - \displaystyle\sum\limits_{j=0}^\infty (1+z)^n$ .

So we have $\displaystyle - \displaystyle\sum\limits_{j=0}^\infty (1+z)^n * (1+z)^{-1} = - \displaystyle\sum\limits_{j=0}^\infty (1+z)^{n-1} = - \displaystyle\sum\limits_{j=-1}^\infty (1+z)^n$. Hurray, case closed!