The Analytic functions are defined as $f$ is analytic on the domain $D$ if for any $x_0$ in $D$, one can write $$f(x)=\sum_{n=0}^{\infty}a_n(x-x_0)^n$$ and the series is convergent to $f(x)$ for $x$ in the neighbourhood of $x_0$.
Equivalently, it's Taylor Series: $$f(x)=\sum_{n=0}^{\infty}\frac{f^n(x_0)}{n!}(x-x_0)^n$$ converges to $f(x)$ for $x$ in the neighbourhood of $x_0$ (in the mean square sense).
This is more of copy from Wikipedia.
I have a few trivial questions
Q. First is it different from saying, a function is analytic if it can be well approximated with Taylor Series (I mean is not for some $x$ in the neighborhood of $x_0$ redundant)?
Q. Second, what does the note in bracket "(in the mean square sense)" mean? Is mean square sense=reference to euclidean metric?
Q. From this definition how can I conclude that $e^{(-1/z)},z\in \mathbb{C}$ is not analytic?
Edit Let's define $z \neq 0$ . Then is it analytic or not?
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$\begingroup$As far as the third question is concerned: consider the closely related function defined on $\mathbf{R}$ by $f(x) = 0$ for $x\leqslant 0$ and $f(x) = \exp(-1/x)$ otherwise is not analytic in $x=0$. For clearly in every neighbourhood of $0$ there exists points in $\mathbf{R}$ such that $f(x)\neq 0$, however it can be shown by induction that $(D^n f)(0) = 0$ so that the Taylor series expansion around 0 vanishes and therefore cannot represent $f$.
$\endgroup$ 0 $\begingroup$Assuming that $z \neq 0$ then the function is holomorphic. There are several ways to prove this. The obvious one is to consider the Cauchy-Riemann equations:
$$\begin{array}{ccc} U(x,y) &=& e^{-x/(x^2+y^2)}\cos\left(\frac{x}{x^2+y^2}\right) , \\ V(x,y) &=& e^{-y/(x^2+y^2)}\sin\left(\frac{y}{x^2+y^2}\right) . \\ \end{array}$$
A straight-forward application of the chain rule, product rule and quotient rule shows that $U_x = V_y$ and $U_y = -V_x$. Thus $f(x,y) = U(x,y) + iV(x,y)$ is holomorphic for all $z \in \mathbb{C}^{\times}$.
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