TITLE: Are there two conventional definitions of “holomorphic”?
QUESTION [6 upvotes]: In Walter Rudin's Real and Complex Analysis, second edition, on page 213, two definitions are stated.  One of them says the derivative of $f$ at $z_0$ is
$$f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}.\tag 1$$
The other says $f$ is holomorphic in an open set if $f'(z_0)$ exists at every value of $z_0$ in that set.
By that definition, holomorphic doesn't just mean differentiable; it means differentiable at every point in some open set.
He never defines “holomorphic at” a point, but only “holomorphic in” an open set.
Before opening the book this afternoon, just remembering from years ago, I thought what it said was

$f$ is differentiable at $z_0$ if $(1)$ holds; and
$f$ is holomorphic at $z_0$ if the derivative exists at every point in some open neighborhood of $z_0$.

By that definition $f(z)=|z|^2$ would be differentiable at $0$ but not holomorphic at $0$.
Thus differentiability at an isolated point would be weaker than holomorphy at that point.
QUESTION: Do both conventions exist?
I take “$f$ is analytic at $z_0$” to mean $f$ has a convergent power series expansion at $z_0$, which actually converges to the right thing, the value of $f$, in the interior of its circle of convergence. Differentiability at every point in some open neighborhood of $z_0$ is enough to prove analyticity, but differentiability at $z_0$ is not. By one of the conventions above, one can say a function holomorphic at a point is analytic at that point; by the other one cannot.

REPLY [3 votes]: I am writing this like a answer, but only because for me to add a comment here I have t have 50 reputation (but I don't have still).
So, Michael Hardy, I was with a similar doubt. Because in Churchill's Complex variables and its applications, there's a section where he define an analytical function just like a function which derivative exists in a point $z_{0}$ and in the points that are a neighborhood of the point $z_{0}$.
In the case of the function $f(z)={\vert}z{\vert}^{2}$, this function has a derivative defined in the point $z=0$, but in the $z=0$ neighborhood we cannot define the derivative (because the lateral limit does not exist, so the derivative does not exists too). 
But there is another case:  the function $g(z)={\frac{1}{z}}$. It's derivative is not defined for the point $z=0$, but it exists to the neighbor points of $z=0$. In this case, $z=0$ is a singular point of the function.
Note: I would like to write this like an comment (in fact, this is not the answer!), but my S.E. points are not enough to post a comment.