This question is from Pg 457 of Ponnusamy and Silvermann's Complex analyis book.
How to prove that set of regular points of an analytic function is open ?
$z_1$ is called regular point wrt analytic function f(z) if for $z_1$ there exists a curve such that function element (f,D) can be analytically from the point in D to point $z_1$.
If $z_1$ is assumed to be analytic then it doesn't means that i can draw an open 2 -dimensional ball around $z_1$ so that each point inside the ball is also regular point.
I think I need to assume atleast 1 sigular point in that open ball but I am unable to find any contradiction.Can you please tell which result should i use?
In case it is helpful, I have proved that the set of singular points of an analytic function form a closed set.
Singular points are those points on the boundary of Domain of function element which are not regular points.
The claim follows from the fact that a domain in the book is defined as an open (non-empty) connected subset of the complex plane.
Using the notation in the book: Let $(f,D)$ be a function element (i.e. a domain $D$ and an analytic function $f$ on $D$). A point $z_1$ is a regular point if there is a chain of function elements $(f_k,D_k)$, $k=0,...,n$ and a continuous curve $\gamma:[0,1]\mapsto {\Bbb C}$ with the following properties:
(1) For $0\leq k<n$ we have $D_k\cap D_{k+1} \neq \emptyset$ and ${f_k}_{|D_k\cap D_{k+1}} = {f_{k+1}}_{|D_k\cap D_{k+1}}$.
(2) $z_0=\gamma(0)\in D\cap D_0$, $\gamma(1)=z_1\in D_n$ and $\gamma([0,1])\subset D_0\cup ... \cup D_n$.
Now, since $D_n$ is open there is $r>0$ such that $B(z_1,r)\subset D_n$. If $w\in B(z_1,r)$ then $f_n$ is analytic in a neighborhood of $w$ and by concatenating the path $\gamma$ and the line-segment $[z_1;w]$ joining the two points we construct a new path $\tilde{\gamma}$ for which all the above properties still hold with $w$ now being the end-point. Thus, every point in $B(z_1,r)$ is a regular point.