Holomorphic equivalent to analytic A holomorphic function is differentiable everywhere and satisfies Cauchy-Riemann condition.
Prove that a function is holomorphic if and only if it's analytic? 
I have no idea how to prove this.
Thanks...
 A: Let $G$ be a domain and $f: G \to \mathbb{C}$ be a function who is differentiable everywhere in $G$ and satisfies Cauchy-Riemann equations, that is according to your definition $f$ is holomorphic in $G$. Also let $a \in G$ and $R>0$ such that $D(a,R)=\{z \in G : | z-a |< R \}\subset G$, then $f$ is holomorphic in $D(a,R)$. Now take any $r$ such that $0<r<R$ and define the path $\gamma(t)=a+re^{it}$ for $t \in [0, 2\pi]$. If $\zeta \in \partial{D(a,r)}=\gamma([0,2\pi])$ and $z \in D(a,r)$ then clearly 
$$
\left| \frac{z-a}{\zeta-a} \right|<1
$$
which gives that 
$$
\frac{1}{\zeta-z} = \frac{1}{\zeta-a}\left( \frac{1}{1-\frac{z-a}{\zeta-a}}\right) = \frac{1}{\zeta-a}\sum_{n=0}^{\infty}\left(\frac{z-a}{\zeta-a}\right)^n = \sum_{n=0}^{\infty}\frac{(z-a)^n}{(\zeta-a)^{n+1}} \ \ (1)
$$
Furthermore, if $M=\max\{|f(\zeta)|: |\zeta-a|=r\}$, then since $|z-a|/r<1$ and
$$
|f(\zeta)|\frac{|z-a|^n}{|\zeta-a|^{n+1}} \leq \frac{M}{r}\left(\frac{|z-a|}{r}\right)^n
$$
the $M$-test of Weierstrass gives that the series (1) times $f(\zeta)$ converges uniformly to $f(\zeta)/(\zeta-z)$. That is 
$$
\frac{f(\zeta)}{\zeta-z}= \lim_{k \to \infty} f(\zeta)\sum_{n=0}^{k}\frac{(z-a)^n}{(\zeta-a)^{n+1}} \ \ (2) \ \ \ \ \ \ \ \text{where the convergence is uniform    (3)}
$$
Then, finally by the C.I.F we have that for all $z \in D(a,r)$,
\begin{align*}
f(z) & \overset{CIF}{=} \frac{1}{2\pi i} \int_{\gamma}\frac{f(\zeta)}{\zeta-z} d\zeta \\
& \overset{(2)}{=} \frac{1}{2\pi i} \int_{\gamma}\lim_{k \to \infty} f(\zeta)\sum_{n=0}^{k}\frac{(z-a)^n}{(\zeta-a)^{n+1}}d\zeta \\
&\overset{(3)}{=}  \lim_{k \to \infty} \sum_{n=0}^{k}\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-a)^{n+1}}d\zeta\right) (z-a)^n\\
&= \sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-a)^{n+1}}d\zeta\right) (z-a)^n\\
&= \sum_{n=0}^{\infty}a_n(\gamma) (z-a)^n
\end{align*}
where 
$$
a_n(\gamma)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-a)^{n+1}}d \zeta
$$
Moreover, since you must already know that if $f(z)= \sum_{n=0}^{\infty}a_n (z-a)^n$ then $a_n=f^{(n)}(a)/n!$, then $a_n(\gamma)=a_n=f^{(n)}(a)/n!$, and you have now that the coefficients $a_n$ don't depend on $\gamma$, thus they are independent of $r$. Then we now have that indeed $f$ has a power expansion of the form $f(z)= \sum_{n=0}^{\infty}a_n (z-a)^n$, and this gives that $f$ is analytic $\blacksquare$.
Also as a corollary we get that for all $0<r<R$, 
$$
\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{(\zeta-a)^{n+1}}d \zeta = \frac{f^{(n)}(a)}{n!}
$$
