Let $a\in\mathbb{C}$ and $k\in\mathbb{N}$, we wish to compute the Laurent Series for the function $$ f(z)=\frac{1}{(z-a)^k} $$ about $z=0$ (NOT $z=a$). So there should be two Laurent Series which are valid for $|z|<|a|$ and $|z|>|a|$. Clearly if $a=0$ then this is already a Laurent Series. But what about $a\ne 0$? I can do $$ \frac{1}{z-a}, $$ but not quite sure how to do $(z-a)^{-k}$. Could somebody give me some clue about how to do this or give me some general methods to do this kind of question?
First for the case $|z|<|a|$, $f(z)$ is holomorphic, then it has a Taylor Series $$ f(z)=\frac{1}{(z-a)^k}=\sum^{\infty}_{n=0}a_nz^n. $$ Then $$ a_n=\frac{f^{(n)}(0)}{n!}=\frac{(-1)^n(k+n-1)!}{n!(k-1)!(0-a)^{k+n}}=\frac{(-1)^k}{a^{k+n}}\binom{n+k-1}{k-1}. $$ In particular, if $a=1$, then for $|z|<1$ $$ \frac{1}{(z-1)^k}=(-1)^k\sum^{\infty}_{n=0}\binom{n+k-1}{k-1}z^n. $$
Now if $|z|>|a|$, then $\left|\frac{1}{z}\right|<|a|$, we write $$ f(z)=\frac{1}{(z-a)^k}=\frac{1}{(-1)^k(a-z)^k}=\frac{1}{(-1)^kz^k(\frac{a}{z}-1)^k}. $$ From above, we know that $\left|\frac{a}{z}\right|<1$, then $$ \frac{1}{(\frac{a}{z}-1)^k}=(-1)^k\sum^{\infty}_{n=0}\binom{n+k-1}{k-1}\left(\frac{a}{z}\right)^n. $$ Then for $|z|>|a|$, we have $$ f(z)=\sum^{\infty}_{n=0}\binom{n+k-1}{k-1}\frac{a^n}{z^{n+k}}=\sum^{\infty}_{n=k}\binom{n-1}{k-1}a^{n-k}z^{-n}. $$
Observe that $$ \frac{1}{(z-a)^{k+1}} = \frac{(-1)^k}{k!}\frac{d^{k}}{dz^{k}}\left(\frac{1}{z-a}\right) $$ and use the fact that you can differentiate the power series of $\frac{1}{1-x}$ term by term.