What is given to prove looks like the definition Suppose  that  $f(z)$   is  analytic  at  $z_0$ . With power series expansion $$f(z)=\sum a_k(z-z_0)^k$$ 
Then  the  radius  of  convergence  of  the  power  series  is  the  largest  number  $R$  such  that  $f(z)$  extends  to  be  analytic  on  the  disk $\{|z-z_0|\lt R\}$ 
How  can  I  prove this  corollary $?$  It  is  corollary  to  the  theorem  that  states that  analytic  function  $f(z)$  has  power  series  expansion as  above  and  $$a_k={f^k(z_0)\over {k!}}$$ . How  is  that  theorem  leading  to  this  corollary  $?$ 
What  am  I  missing $?$
.Please  help  me to  prove it.Thanks. 
 A: The subtlety is that $f$ being analytic at $z_0$ means that $f(z) = \sum a_k (z - z_0)^k$ in some disk centered at $z_0$.  The radius of that disk certainly cannot be larger than $R$, but a priori it could be smaller.  It could be that $\sum a_k (z - z_0)^k$ converges for some values of $z$ but does not equal to $f(z)$ at those points.  The problem is asking you to prove that this hypothetical situation actually can never occur.
A: The question is a little imprecise (and this answer a little pedantic). An essential issue here is connectedness.
For example, let $f(z) = 0$ for $|z|<1$ and $f(z) = 1$ for $|z|>1$. Then
$f$ is analytic everywhere except $|z|=1$ and the power series representation
of $f$ at $z=0$ is, of course, $z \mapsto\sum_k 0 z^k$, which has an infinite
radius of convergence, but clearly $f$ cannot be extended to be analytic on the whole plane.
A better statement might be that if $D$ is an open connected set, and
$f:D \to \mathbb{C}$ is analytic on $D$, with $z_0 \in D$ and $a_k,R>0$ as
above, then there is an (unique) analytic extension of $f$ to $\tilde{f}:D \cup B(z_0,R) \to \mathbb{C}$ such that $f(z) = \tilde{f}(z)$ for $z \in D$.
Furthermore, if $R'>R$ there is no analytic extension to $D \cup B(z_0,R')$.
The proof of this statement relies on the following fact: Suppose $C$ is an open connected set and $a,b:C \to \mathbb{C}$ are analytic on $C$, and furthermore, there is
a set $L$ which has a limit point in $C$ such that $a(z) = b(z)$ for all $z \in L$. Then $a(z)=b(z)$ for all $z \in C$.
Let $g(z) = \sum_k a_k (z-z_0)^k$. Then $g$ is an analytic function on $B(z_0,R)$. Furthermore,
if $g$ is analytic on $B(z_0,R')$ then $R' \le R$ (if $R'>R$, this would
contradict the definition of $R$).
Now suppose $f$ is analytic at $z_0$, and $U \subset D\cap B(z_0,R)$ is an open set containing $z_0$. Then
$f(z) = \sum_k a_k (z-z_0)^k$ for $z \in U$. In particular, $f(z) = g(z)$ for $z \in U$ (since $z_0$ is a limit point of $U$), and hence
$f(z) = g(z)$ for $z \in D\cap B(z_0,R)$.
Now define $\tilde{f}(z) = \begin{cases} g(z) , & z \in B(z_0,R) \\
f(z), & z \in D \setminus B(z_0,R) \end{cases}$. It is straightforward to check
that $\tilde{f}$ is analytic and that $\tilde{f}(z) = f(z)$ for $z \in D$.
If $f$ could be analytically extended to $D \cup B(z_0,R')$ with $R'>R$, then
the expansion $g(z)$ would be valid on $B(z_0,R')$, contradicting the
definition of $R$.
