Prove that the zeros of an analytic function are finite and isolated Let us assume that the zeros of $f = \{Z_1,\ldots,Z_n,a\}$ are infinite and converge towards $a$.
The book which I am reading says that any neighborhood of $a$ will contain infinite zeros. Since $f$ is continuous, $f$ must be identically zero in this neighborhood. .....(A)
Then as the neighborhood grows, it will be identically $0$ in that neighborhood too until it engulfs the whole of this region. Hence, the function becomes identically zero in the whole region.
I don't quite understand (A). I am confused over (A) with this analogy. Consider $f(z) = z-1$. This has a zero at $z=1$. Since $f$ is continuous in the $z$ plane, this means that as per (A) points in the neighborhood of $z =1$ should also have mapped value $= 0$.
I am confused; where could I be making a mistake?
Thanks
 A: I think that problem is that continuity is not enough. (Although I can't think of a counter-example off-hand). What you really need is for the function to be holomorphic in some neighbourhood of it's zeroes.
I know of the related result (known as the "principle of isolated zeroes"):
Suppose $f:B_a(r)\rightarrow \mathbb{C}$ is a non-zero holomorphic function on $B_a(r)$with $f(a) = 0$. Then $\exists 0 <\rho<r$ such that $\forall z \in B_a(\rho)\text{ \ }\{a\}, f(z) \not = 0$.
To prove this, we need Taylor's theorem. This says that on $B_a(r)$, $f$ can be written as
$$f(z) =\sum_{n=0}^\infty a_n(z-a)^n$$
 If $N$ is the largest integer such that $a_n = 0$ for all $n \leq N$ then we can write $f(z) = (z-a)^Ng(z)$ where $g$ is a holomorphic function and $g(a)$ is not zero. By the continuity of $g$ there is some open ball around $a$ (call it $B_a(\rho)$) where $g$ is not zero, so since $(z-a)^N$ is non-zero in $B_a(\rho)\text{ \ }\{a\}$, $f$ is also non-zero there.
A: The set of Zeros of an analytic function may be countably infinite. For example $f(z)=\sin\left(\frac1z\right)$ is analytic on $C\setminus\{0\}$. It is true that The set of Zeros of an analytic function at most countably infinite. 
A: The statement is wrong the way you quoted it. Take $f \,:\, \mathbb{R} \to \mathbb{R}$, $$
  f(x) = \begin{cases}
    x\sin \frac{1}{x} &\textrm{if } x > 0 \\
    -x &\textrm{if } x \leq 0 \text{.}
  \end{cases}
$$
Since $\lim_{x\to 0^+} f(x) = 0$ (and obviously also $\lim_{x\to 0^-} f(x) = 0$), that function is continuous. It has zeros at $$
  \left\{\frac{1}{n\pi} \,:\, n \in \mathbb{N}\right\} \text{,}
$$
which obviously have $0$ as a cluster point. Yet $f$ isn't identically zero on any neighbourhood of $0$.
For analytic functions, however, you also have that they are $C^\infty$. Your probably have to use that. Note that that only requiring $C^\infty$ isn't enough. Take for example $$
  f(x) = \begin{cases}
    e^{-\frac{1}{x}} \sin \frac{1}{x} &\textrm{if } x > 0 \\
    -x &\textrm{if } x \leq 0 \text{.}
  \end{cases}
$$
That $e^{-\frac{1}{x}}$ should damp things strong enough to ensure the existance of all derivatives of $f$ at zero, i.e. $f$ should be $C^\infty$, yet the theorem still doesn't hold since $f$ still isn't identically zero on a neighbourhood of zero.

So from the above it's clear the we need to use that $f$ is analytic to prove the theorem. We'll wlog assume that $a = 0$ (simply look at $f'(x)=f(x+a)$ if it isn't). Let $(x_n)$ be a sequence of zeros of $f$, with $x_n \to 0$. Since $f$ is analytic at $0$, there's a $r \geq 0$ such that for all $|x| < r$ we have $$
  f(x) = \sum_{k=0}^\infty c_k x^k \text{.}
$$
Let's further wlog assume that $|x_n| < r$ for all $n$ and that $|x_{n+1}| < |x_n|$ (Simply remove those $x_n$ which violate one of the conditions. Since the sequence converges to 0, infinitely many $x_n$ will remain). From the continuity of $f$ it follows that $f(0)=0$, i.e. that $c_0 = 0$. For $f(x_n)$ to be zero, it must hold that \begin{align}
  &\sum_{k=0}^\infty c_k x_n^k = 0
  \implies
  \underbrace{c_0}_{=0} + c_1 x_n = -x_n^2\sum_{k=2}^\infty c_k x_n^{k-2}
  \implies
  c_1 = -x_n\sum_{k=2}^\infty c_k x_n^{k-2} \\
  \implies
  &|c_1| \leq |x_n|\underbrace{\sum_{k=2}^\infty |c_k| |x_n|^{k-2}}_{:=M_n} \text{.}
\end{align}
(The last step uses that the series is absolutely convergent, which you have for series expansions of analytic functions)
Now observe that since $|x_{n+1} < x_n|$ you have $M_{n+1} \leq M_n$, and hence for arbitrary $n$ that $|c_1| \leq |x_n|M_n$. But since $x_n \to 0$, it follows that $c_1=0$.
Knowing that $c_1=0$, you can use the same method to show that, in fact $c_2=0$, and from that $c_3=$, and so on. You thus get that all the $c_n$ are zero, and hence that $f$ is identically zero on the ball with radius $r$ around $0$.
