# Help with a Laurent expansion

#### Pfeffermynz

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.

#### chisigma

MHF Hall of Honor
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

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

#### Pfeffermynz

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!