There exists an deleted neighborhood around $0$ so thag a non-zero analytic function is no-where $0$ This is obvious if we use Identity theorem. But I can't use that since it is real analysis exercise. All I can use there exists series expansion for the function around $0$.
Since non-zero analytic function not all coefficient is $0$. So we need to find $r$ so that $0<|z| \leq r$ such that $\sum a_n z^n$ is never zero. We need to prove only using the techniques of sequence and series. How to proceed? Any help please?
 A: I will assume that the Taylor series of $f$ centered at $0$ is not the null series; otherwise, the statement is trivially false.
If $f(0)\ne0$, it follows from the continuity of $f$ that there is a neighborhood of $0$ at which there are no zeros of $f$. If you want to avoid assuming that $f$ is continuous, you can do it as follows: near $0$, you have\begin{align}f(x)-f(0)&=\left(\sum_{n=0}^\infty a_nx^n\right)-a_0\\&=\sum_{n=1}^\infty a_nx^n\\&=x\sum_{n=0}^\infty a_{n+1}x^n.\end{align}Now, let $\rho$ be the radius of convergence of the series $\sum_{n=0}^\infty a_nx^n$ and take $r\in(0,\rho)$. If $|x|<r$, it follows from what was done above that$$\bigl|f(x)-f(0)\bigr|<|x|\sum_{n=0}^\infty|a_{n+1}|r^n,$$and so, if$$|x|<\frac{|f(0)|}{\sum_{n=0}^\infty|a_{n+1}|r^n},$$then $\bigl|f(x)-f(0)\bigr|<\bigl|f(0)\bigr|$, and therefore $f(x)\in\bigl(0,2f(0)\bigr)$; in particular, $f(x)\ne0$.
And if $f(0)=0$, then let $k$ be the smallest natural number such that $a_k\ne0$. Then$$f(x)=a_kx^k\left(1+\frac{a_{k+1}}{a_k}x+\frac{a_{k+2}}{a_k}x^2+\cdots\right).$$If$$g(x)=1+\frac{a_{k+1}}{a_k}x+\frac{a_{k+2}}{a_k}x^2+\cdots,$$then $g(0)=1\ne0$ and therefore $g$ has no zeros near $0$ (as above) And, of course, $a_kx^k=0\iff x=0$. So, near $0$, the only zero of $f$ is $f$ itself.
