Finding the $n$th Taylor coefficient of $g(z)=\frac{z}{(z-b)^2}$ centered at $a$ (where $a=2-\sqrt{3}$ and $b=2+\sqrt{3}$? I've introduced $a$ and $b$ in order to simplify the notation : $a=2-\sqrt{3}$ and $b=2+\sqrt{3}$.
I'm trying to compute the Taylor Series for $g(z)=\frac{z}{(z-b)^2}$ centered at $a$. I denote the $n$th Taylor coefficient as $c_n$, then
$$c_n= \frac{1}{2\pi i} \int_{\gamma(a,r)} \frac{z}{(z-b)²(z-a)^{n+1}}dz$$
where $0<r<2\sqrt{3}$.
Clearly for $n=0$, the computation is easy noting that $\frac{z}{(z-b)^2}$ is holomorphic on and inside $\gamma(a,r)$ with our restriction on $r$. But I can't do anything with this once $n$ is not $0$... Can you help me please? A hint would be great!
Thank you!
 A: Hint.  First, I would note some simple relations between $a$ and $b$, namely
$$a+b=4\ ,\quad b-a=2\sqrt3\ .$$
Second, you want powers of $z-a$ so I would substitute $w=z-a$ and simplify using the above:
$$\eqalign{f(z)
  &=\frac{w+a}{(w+a-b)^2}\cr
  &=\frac{(w+a-b)+b}{(w+a-b)^2}\cr
  &=\frac{1}{w-2\sqrt3}+\frac{b}{(w-2\sqrt3)^2}\ .\cr}$$
Finally, I would do some more simple algebraic manipulations then use the standard series
$$\frac{1}{1-t}=\sum_{n=0}^\infty t^n\ ,\quad
  \frac{1}{(1-t)^2}=\sum_{n=0}^\infty(n+1)t^n\ .$$
I wouldn't use the integral formula at all.  See if you can complete the details.  Don't forget to check that you use these series only within their circle of convergence, and substitute back in terms of $z$ at the end.
A: We have $$\frac{z}{(z-b)^2}=((z-a)+a)((z-a)+(a-b))^{-2}$$
Now use the binomial theorem
$$(X+C)^r=\sum_{k=0}^{\infty}\binom{r}{k}X^kC^{r-k}$$
We get $$((z-a)+(a-b))^{-2}=\sum_{k=0}^{\infty}\binom{-2}{k}(a-b)^{r-k}(z-a)^{k}$$
We can multiply by the other factor $((z-a)+a)$ and we get the series.
A: Hint:
$$h(z)=\frac{1}{z-b}=\sum_{n=0}^\infty a_n(z-a)^n$$
(you can calculate this series easily)
$$h'(z)=\frac{-1}{(z-b)^2}=\cdots$$
$$g(z)=\frac{z}{(z-b)^2}=\frac{(z-a)+a}{(z-b)^2}=\cdots$$
