why $f$ is holomorphic if $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)\, d \zeta}{\zeta - z}$? I'm reading Gong Sheng's Concise Complex Analysis to get some basic understanding.
On $\S 2.4$ page 61 Theorem 2.15 (Hurwitz Theorem) it says

Theorem 2.15 (Hurwitz Theorem) Let $\{f_j\}$ be a sequence of holomorphic functions on $U\subseteq \mathbb C$ that converges uniformly to a function $f$ on 
  every compact subset of $U$. If $f_j$ is never equal to zero on $U$ for all $j$, then 
  $f$ is either identically zero or never equal to zero on $U$.
Proof: For an arbitrary point $z \in U \subseteq \mathbb C$, choose a simple closed curve $\gamma$ in $U$ such that the inside of $\gamma$ contains $z$. 
  Since $f_j$ is holomorphic on $U$, by Cauchy integral formula we have $$f_j(z)=\frac{1}{2\pi i}\int_\gamma \frac{f_j(\zeta)\,d \zeta}{\zeta-z}$$
  Since $\{f_j\}$ converges uniformly on every compact subset of $U$, we have 
  $$\lim\limits_{j\to \infty} f_j(z) = \lim\limits_{j\to \infty} \frac{1}{2\pi j} \int_\gamma \frac{f_j(\zeta)\,d \zeta}{\zeta-z}  = \frac{1}{2\pi j} \int_\gamma \lim\limits_{j\to \infty} \frac{f_j(\zeta)\,d \zeta}{\zeta-z} $$
  It follows that $$f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta) \, d \zeta}{\zeta - z}$$
  Hence, $f(z)$ is a holomorphic function. 
  Similarly, we can prove that $\{f_j'(z)\}$ converges uniformly to $f'(z)$ on every compact subset of $U$. 
If $f(z)$ is not identically zero, then by Theorem 2.13, the zeros of $f$ are  discrete. Let $\gamma$ be a curve that does not pass through these zeros. Then 
  $$\frac{1}{2\pi i} \int_\gamma \frac{f_j'(\zeta)}{f_j(\zeta)\, d \zeta} \to \frac{1}{2\pi i} \int_\gamma \frac{f'(\zeta)}{f(\zeta)\, d \zeta}$$
  as $j\to \infty$.
  By the assumption and Theorem 2.14, we have 
  $$\frac{1}{2\pi i} \int_\gamma \frac{f_j'(\zeta)}{f_j(\zeta)\, d \zeta} = 0$$
  Therefore  $$\frac{1}{2\pi i} \int_\gamma \frac{f'(\zeta)}{f(\zeta)\, d \zeta} = 0$$
  and $f(z)$ has no zero on $U$. 

I wonder why $f(z)$ is a holomorphic function once $f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta) \, d \zeta}{\zeta - z}$?
So far from that book I only know such ways to determine a function is holomorphic


*

*By definition, that is $f(z)$ is holomorphic in $U$, iff $\forall z\in U$, $\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists, here $h \in \mathbb C$. 

*Cauchy-Riemann equation, that is, if $f(z) \in \mathscr L ^1 (U)$ and fulfills the Cauchy-Riemann equation $\frac{\partial f}{\partial x} = -i \frac{\partial f}{\partial y}$, then $f$ is holomorphic on $U$, here $z=x+iy$.

*Power series expansion (Taylor series): $f(z)$ is holomorphic in $U$ iff $f$ has a power series expansion $\forall z \in U$.

*Morera theorem:  If $f(z)$ is continuous on $U$ and 
the integral of $f$ along any rectifiable closed curve is zero, then $f(z)$ is 
holomorphic on $U$. 
It seems none of such 1-4 could get to the conclusion that $f(z)$ is a holomorphic function?
 A: Suppose that $f$ is a continuous function on some domain $D$ of the complex plane, and take a simple closed piecewise smooth curve $\gamma$ inside $D$. Define for $z\in D\setminus |\gamma|$ the function $$F(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$$
If we let $d_z=d(\gamma,z)$ for $z$ inside $\gamma$, one can use the geometric series to show that at each of these points $F$ can be expanded in a convergent powerseries with radius at least $d_z$, whence it follows $F$ is holomorphic. See por example page 209 of Remmert's "Theory of Complex Functions."
A: If it had said $f(z) = \displaystyle\frac{1}{2\pi i} \int_\gamma \frac{g(\zeta) \, d \zeta}{\zeta - z}$ (with two functions, $f$ and $g$, where $g$ is not badly behaved, then I'd be thinking about using Morera's theorem to show $f$ is holomorphic.  But here you have $f$ where I put $g$ above, and also on the left.  Let's see if Morera can still help us.  Let $C$ be a simple closed curve in the interior of the region that $\gamma$ surrounds.  The denominator below will never be $0$ since $C$ is in the interior of that region, and will in fact be bounded away from $0$ because the curve $C$ is compact and so is bounded away from $\gamma$.
\begin{align}
\int_C f(z)\,dz & = \frac{1}{2\pi i} \int_C \int_\gamma \frac{f(\zeta)}{\zeta - z}\,d\zeta  \,dz \overset{\text{ ??? }}= \frac{1}{2\pi i} \int_\gamma \int_C \frac{f(\zeta)}{\zeta - z}\,dz\,d\zeta \\[10pt]
& = \frac{1}{2\pi i} \int_\gamma\left( f(\zeta) \int_C \frac{1}{\zeta - z}\,dz\right)\,d\zeta = \frac 1 {2\pi i} \int_\gamma (f(\zeta)\cdot 0)\,d\zeta = 0.
\end{align}
We can interchange the order of integration in this way if $f$ is not too badly behaved.  If we somehow know $f$ is continuous, that's enough, since we're integrating over a compact set, so it's bounded, and with a bounded function on a set of finite measure we can do that.  Just what hypotheses on $f$, or what other known facts about $f$, do you have?
A: EDIT: this is just an hint, or maybe I am just pointing in the direction I would take to prove this :D
Use point 3. in the following way:
By induction on $n$ you can prove the following formula $$f^{(n)}(z) = \frac{n!}{2 \pi i}\int_{\gamma}\frac{f(\xi)}{(\xi - z)^{n + 1}}\,d\xi,$$
which can be used to get a Taylor expansion for every $z$.
(Good question btw)
