$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$, $\lim a_n = z_0$, then $a_n = 0, \forall n$

Question

I need to show the following:

Consider

$$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n$$

a series with radius of convergence $R>0$. Suppose that there is a non constant sequence $a_n$ such that

$$\lim a_n = z_0$$

and $f(a_n)=0$, then the series is identically null, that is, $a_n=0, \forall n$

First of all, what does $f(a_n)=0$ means? I think it's a typo. Does somebody know what it should mean? Why this result is important?

Could somebody help me in how to prove it?

Answer 1

The other answers are correct, but use some sort of theorem. What you need is

The zeros of a non-constant analytic/holomorphic function are isolated

proof : in the neighborhood of $z_0$ a zero of $f$ $f(z) = f(z_0)+C(z-z_0)^n + o(|z-z_0|^n)=C(z-z_0)^n (1+ o(1))$ for some $n \in \mathbb{N}$ and $C\ne 0$, so that $f(z) \ne 0$ on $0<|z-z_0|<\epsilon$

Now if $f$ is analytic around $z_0$ and has a sequence of zeros $f(a_n)=0$ with $a_n \to z_0$

Then by continuity $f(z_0)=\lim_n f(a_n) = 0$, and if your sequence $a_n$ isn't constant for $n$ large enough, then $z_0$ is a non-isolated zero of $f$. And by the previous theorem, it means that $f$ is identitcally zero.

Answer 2

Is this in a complex analysis course?

The statement $f(a_n) = 0$ means exactly what you think it means. When you plug in the number $a_n$ you happen to get $0$.

Case 1: Suppose the sequence $(a_n)$ has infinitely many values. This means that there is an infinite sequence of distinct inputs that are mapped to 0, and these inputs converge to $z_0$. But $f$ is analytic (has a convergent Taylor series), so that would mean that $f$ is identically $0$. [Just take this subsequence in your limit definitions for derivatives at $z_0$.]

Case 2: The sequence $(a_n)$ takes only finitely many values. This means that the function is of the form $f(z) = p(z) + \sum_{n \geq N} z_0 (z-z_0)^n$ for some polynomial of degree at most $N$ (because eventually all the terms $a_n$ must equal $z_0$). But this actually is perfectly reasonable. Consider say $z_0 = 0$ and say $f(z) = 0 - z + 0 + z^3 + 0 +0+0+ \cdots = z^3 -z.$ this satisfies $f(0) = f(1) = f(-1) =0$ [so $f(a_n) = 0$ for all $n$], and $a_n \to 0 = z_0$.

So the thing seems to be false. But I imagine the intent was not to allow case 2 as a possibility.

Answer 3

First, if $f$ is a power series with radius of convergence $R$, then $f$ is analytic on the ball $B(z_0,R)$. Next, the set $\{a_n: n\in\mathbb{N_0}\}$ is the set of zeroes of $f$, and has an accumulation point, $z_0\in B(z_0,R)$. By the Identity Theorem, $f\equiv 0$.