Expand $(1-z)^{-m}$ in powers of $z$ There is a partial duplicate of this question, but I have an extension of that question.
Expansion of $(1-z)^{-m}$
The problem is "Expand $(1-z)^{-m}$ in powers of $z$ where $m$ is a positive integer".  That post above discusses for $|z|<1$.  I know that the same argument does not work $|z|>1$.  But that does not prove that there is not a different way to do this.  So my question is "Is there a way?"  The reason I feel there might be some way is that this is a question (1.20) from Complex Analysis by Stein and Shakarchi and there is no specification to the value of $z$ or any reference that there might be some restriction we need to put on $z$.
Thanks.
 A: Have your heard of Newton's generalized binomial theorem? http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem.
According to this result, in the case $|z|<1$ the answer is simply
$$ (1-z)^{-m} =\sum_{r=0}^\infty  \binom{-m}{r}(-z)^r $$
Try for yourself using the taylor series expansion at $z=0$. The statement of the theorem is that in general:
$$ (x+y)^s=\sum_{r=0}^\infty \binom{s}{r}x^{s-r}y^r $$.
Note also that the binomial coefficient is well defined for $s\in\mathbb{R}$. There are several proofs available on the internet. By the way, the binomial coefficient for $s$ any real numbers is just
$$ \binom{s}{r} = \frac{s(s-1)...(s-r+1)}{r!}$$
REMARK: When we say $|z|<1$ it doesn't mean that the function will only have an expression in power series near $0$, it actually means that THE GIVEN particular series expansion will be only valid for these region. BUT YOU CAN CHOOSE ANY OTHER REGION! Do this simply by taking the Taylor expansion around some other value, say $z_0=3$, then your formula will be valid for the numbers $2<z<4$ (if $z$ is real, otherwise for values such that $|z-3|<1$). 
To do this in a faster way for the particular function you are giving, you can do do a simple trick using again Newton's generalized theorem,
$$  (1-z)^{-m} = ((1-z_0)+(z_0-z))^{-m} = \sum_{r=1}^\infty \binom{-m}{r}(1-z_0)^{-m-r}(z_0-z)^r = \sum_{r=0}^\infty \binom{-m}{r}(-1)^{-m-r}(1-z_0)^{-m-r}(z-z_0)^r $$
This expansion will be valid for the values of $z$ such that $|z-z_0|<1$. Hope this helps.
