Hey folks! I was trying to find the Laurent series of f=1/(z^2+z) in the region 1<|z-1|<2.

f can be expressed as
1/(w+1) - 1/(w+2), where w=z -1

f=1/(1-(-w)) - (1/2)(1/(1-(-w/2))

Anyway, at first I tried making 1/(1-(-w))= the sum from 0 to infinity of (-w)
^n. However, that's only valid when |w|<1, which is not the case here, so I got a wrong answer.

Then I looked for some examples on the internet and found the following modified geometric series

1/(1-z)= -(1/z
^0 + 1/z^1 + 1/z^2+...), for |z|>1.

Using this formula I got the right answer. My question now is, how is that formula derived?

2. ## Re: Question about series

Suppose $\displaystyle |z|>1$. Then $\displaystyle \dfrac{1}{1-z} = \dfrac{\tfrac{1}{z}}{\tfrac{1}{z}-1}} = -\dfrac{1}{z}\dfrac{1}{1 - \tfrac{1}{z}}$. Since $\displaystyle |z|>1, \left|\dfrac{1}{z}\right|<1$.

3. ## Re: Question about series

Originally Posted by SlipEternal
Suppose $\displaystyle |z|>1$. Then $\displaystyle \dfrac{1}{1-z} = \dfrac{\tfrac{1}{z}}{\tfrac{1}{z}-1}} = -\dfrac{1}{z}\dfrac{1}{1 - \tfrac{1}{z}}$. Since $\displaystyle |z|>1, \left|\dfrac{1}{z}\right|<1$.
Of course. Thank you!