'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.