Laurent series of a function

Question

I want to calculate the Laurent series expansion of the function ${z-1 \over z+1}$ Centered at $z=0$. I'd really appreciate some hints.

Answer 1

Just for giggles (such is my exciting life...):

Let $f$ be the function and note that $(z+1)f(z) = z-1$, so we get $(z+1) f'(z) + f(z) = 1$, and $(z+1) f^{(2)}(z) + 2f^{(1)}(z) = 0$. It is not hard to show that for $k \ge1$ we have $(z+1) f^{(k+1)}(z) + kf^{(k)}(z) = 0$, and so we conclude that $f(0) = -1, f'(0) = 2$ and $f^{(k)}(0) = 2 k! (-1)^{k+1} $ for $k \ge 1$, hence if $f(z) = \sum_k a_n z^k$, we have $a_0 = -1$ and $a_k = {f^{(k)}(0) \over k!} =2 (-1)^{k+1}$ for $k \ge 1$.

Answer 2

You just write

$${z-1\over z+1} = {(z+1)-2\over z+1} = 1 + {-2\over z+1}$$

now expand the latter in geometric series to get

$$1-2\sum_{n=0}^\infty (-1)^n z^n= -1 +\sum_{n=1}^\infty (-1)^{n+1}2z^n$$