extract Catalan numbers from generating function via residues Consider the catalan numbers $C_n=\frac{1}{n+1}\binom{2n}{n}$, which have the following generating function:
$$C(x)=\sum_{n=0}^{\infty}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}.$$
I am well aware of the usual proofs of this identity: for example, one can write $C(x)$ as the root of a quadratic equation via combinatorial identities, or one can Taylor-expand $\sqrt{1-4x}$.
I am wondering if there is a way to start from $C(x)=\frac{1-\sqrt{1-4x}}{2x}$ and extract the Taylor coefficients via residue calculus. That is, if there is a direct way to evaluate the integral
$$\oint_{C}\frac{1-\sqrt{1-4z}}{2z^{n+1}}dz$$
for some appropriately chosen $C$.
 A: The solution could have been much shorter, but I prefer details to concision.
One wants to show that
$$C_n=\frac1{2\pi i}\oint_{C}\frac{1-\sqrt{1-4z}}{2z}\frac{dz}{z^{n+1}}$$
It is easy to see 
$$\oint_{C}\frac{1-\sqrt{1-4z}}{2z}\frac{dz}{z^{n+1}}=-\frac12\oint_{C}\frac{\sqrt{1-4z}}{z^{n+2}}dz$$

Definitions
We will take the branch cut of $\sqrt{1-4z}$ on $\mathbb R_{\ge1/4}$, i.e.
$$\sqrt{1-4z}=\exp\bigg[\frac{\ln|1-4z|+i\arg(1-4z)}2\bigg]\qquad{\arg(1-4z)\in[-\pi,\pi)}$$
I will show how to evaluate
$$I_n:=\oint_C\sqrt{1-4z}~~ z^{-n-2}dz$$ directly.

Decomposing the contour integral
Take $C$ to be the keyhole contour centered at $1/4$ which avoids the branch cut.
The two integrals, one around branch point and one along the infinitely large circle, vanish.
The integral above the real axis $I^+$ is (using the parametrization $z=\frac14+te^{i\theta}$, under the limit $\theta\to0^+$)
$$\begin{align}
I^+
&=\lim_{\theta\to0^+}\int^\infty_0\frac{\sqrt{1-4(1/4+te^{i\theta})}} {(1/4+te^{i\theta})^{n+2}}e^{i\theta}dt \\
&=\int^\infty_0\frac{\lim_{\theta\to0^+}\sqrt{4te^{i(\theta-\pi)}}} {(1/4+t)^{n+2}}dt \\
&=-2i\int^\infty_0\frac{\sqrt{t}}{(1/4+t)^{n+2}}dt \\
\end{align}
$$
The integral below the real axis $I^-$ is (using the parametrization $z=\frac14+te^{i\theta}$, under the limit $\theta\to2\pi^-$)
$$\begin{align}
I^-
&=\lim_{\theta\to2\pi^-}\int_\infty^0\frac{\sqrt{1-4(1/4+te^{i\theta})}}{(1/4+te^{i\theta})^{n+2}}e^{i\theta}dt \\
&=\int_\infty^0\frac{\lim_{\theta\to2\pi^-}\sqrt{4te^{i(\theta-\pi)}}}{(1/4+t)^{n+2}}dt \\
&=-2i\int^\infty_0\frac{\sqrt{t}}{(1/4+t)^{n+2}}dt=I^+
\end{align}
$$
Therefore, $I_n\equiv I^+ + I^-=2I^+=-4i\displaystyle{\int^\infty_0\frac{\sqrt{t}}{(1/4+t)^{n+2}}dt}$.

Evaluating $I_n$ directly
By the substitution $u=\frac1{1+4t}$,
$$\begin{align}
\frac1{-4i}I_n
&=\int^\infty_0\frac{\sqrt{t}}{(1/4+t)^{n+2}}dt \\
&=2^{2n+1}\int^1_0\sqrt{1-u}~~u^{n-1/2}du \\
&=2^{2n+1}B\left(\frac32,n+\frac12\right)\\
&=2^{2n+1}\cdot\Gamma(3/2)\cdot\Gamma(n+1/2)\cdot\frac1{\Gamma(n+2)}\\
&=2^{2n+1}\cdot\frac{\sqrt\pi}{2}\cdot\frac{(2n)!}{2^{2n}n!}\sqrt\pi\cdot\frac1{n+1}\frac1{n!}\\
&=\pi\cdot\frac1{n+1}\binom{2n}{n}\\
I_n&=\frac{-4\pi i}{n+1}\binom{2n}{n}\\
\end{align}
$$

Assembling
$$\oint_{C}\frac{1-\sqrt{1-4z}}{2z^{n+2}}dz=-\frac12\oint_{C}\frac{\sqrt{1-4z}}{z^{n+2}}dz=-\frac12I_n$$
$$=-\frac12\cdot\frac{-4\pi i}{n+1}\binom{2n}{n}=2\pi i\frac1{n+1}\binom{2n}{n}$$
Thus,
$$\color{red}{\oint_{C}\frac{1-\sqrt{1-4z}}{2z}\frac{dz}{z^{n+1}}=2\pi i~C_n}$$ as expected.
A: With the functional equation
$$C(z) = 1 + z C(z)^2$$
and the solution
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}$$
we may choose the branch cut of the square root to be $(1/4, +\infty)$
so  that we  have  an analytic  function in  the  neighborhood of  the
origin and the Cauchy Coefficient Formula applies.
We have
$$n C_n = [z^{n-1}] C'(z) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} C'(z) \; dz.$$
With the substitution $C(z) = w$ we get that $z = (w-1)/w^2$
so that
$$n C_n =
\frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n}}{(w-1)^n} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{1}{(w-1)^n}
\sum_{q=0}^{2n} {2n\choose q} (w-1)^q \; dw
\\ = {2n\choose n-1}.$$
We thus have
$$C_n = \frac{1}{n} {2n\choose n-1}
= \frac{1}{n+1} {2n\choose n}$$
as required. This was  for $n\ge 1$ and we see  that this last formula
also yields the correct value for $n=0.$  Here we have made use of the
fact that $C(z) =  1 + z + \cdots$ so the image  contour of the circle
$|z|=\epsilon$ can  be deformed  to a  circle $|w-1|=\gamma$  making a
single  turn.  The  chosen  branch  cut in  fact  corresponds  to  the
principal branch of the logarithm with argument $(-\pi, \pi].$
