Help with a Laurent expansion

May 2010
2
0
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
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
May 2010
2
0
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!