Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\begin{equation}
\sqrt{1 - 4z} = \sum_{n \geq 0}\binom{1/2}{n}(-4 x)^n.
\end{equation}
and therefore
\begin{align*}
  f(z) = \frac{1 - \sqrt{1 - 4z}}{2} &= \frac{\sum_{n \geq 1}\binom{1/2}{n}(-4)^n z^n}{2} \\
  &= \sum_{n \geq 1}\binom{1/2}{n}(-1)^n 2^{n} z^n \\
  &= \sum_{n \geq 1}\binom{1/2}{n}(-2)^n z^n \\
  &= \sum_{n \geq 0}a_n z^n
\end{align*}
with coefficients $a_0 = 0$ and $a_n = \binom{1/2}{n}(-2)^n$.
I wanted to use cauchys integral formula for
$g(z) = f(z)^r$ to extract the $n$ coefficient
but then I get
\begin{align}
  \left( \frac{1 - \sqrt{1 - 4z}}{2}\right)^r &= \frac{1}{2^r} \left(1 - \sqrt{ 1 - 4z} \right)^r \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sqrt{1 - 4z}^m \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m (1 - 4z)^{m/2} \\
  &= \frac{1}{2^r} \sum^{r}_{m=0}\binom{r}{m}(-1)^m \sum_{k \geq 0}\binom{m/2}{k}(-1)^k 4^k z^k \\
  &= \frac{1}{2^r}\sum^{r}_{m=0} \sum_{k \geq 0}\binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k.
\end{align}
Therefore I should have
\begin{align*}
  [z^n] (f(z))^r &= [z^n] \sum^{r}_{m=0} \sum_{k \geq 0}\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k \\
  &= \sum^{r}_{m=0}[z^n] \sum_{k \geq 0}{\frac{1}{2^r} \binom{r}{m} \binom{m/2}{k} (-1)^{m+k} 4^k z^k} \\
  &= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^{m + n} 4^n \\
  &= \sum^{r}_{m=0}\frac{1}{2^r}\binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
  &= \frac{1}{2^r} \sum^{r}_{m=0} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n \\
  &= \frac{1}{2^r} \sum^{r}_{m=2 n} \binom{r}{m} \binom{m/2}{n} (-1)^m (-4)^n 
\end{align*}
For $r \geq 2n$ otherwise the coefficient vanishes.
I would be thankful if anyone can give me a hint.

We seek
$$[z^k] \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
where $k\ge r$ and we get zero otherwise since $\frac{1-\sqrt{1-4z}}{2}
= z + \cdots.$
Using the residue operator this is
$$\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{k+1}} \left(\frac{1-\sqrt{1-4z}}{2}\right)^r.$$
Now put $\frac{1-\sqrt{1-4z}}{2} = w$ so that $z = w(1-w)$ and $dz\; =
(1-2w) \; dw$ to get
$$\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k+1} (1-w)^{k+1}} w^r (1-2w)
= \; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k-r+1} (1-w)^{k+1}} (1-2w) 
\\ = {2k-r\choose k} - 2 {2k-r-1\choose k}
= {2k-r\choose k} - 2 \frac{k-r}{2k-r} {2k-r\choose k}
\\ = \frac{r}{2k-r} {2k-r\choose k}.$$
Following the other post we may write
$$\sum_{k\ge r} \frac{r}{2k-r} {2k-r\choose k} z^k
= z^r \sum_{k\ge 0} \frac{r}{2k+r} {2k+r\choose k} z^k.$$
The residue operator returns zero when $k\lt r.$
