uniform convergence for a power series

**jbpellerin** · September 28th 2008, 06:28 PM

Let $\sum_{n=0}^\infty a_n(z-z_o)^n$ be a complex power series that converges uniformly on $\mathbb{C}$ .
Show that there is ${N}\epsilon \mathbb{N}$ such that $a_n = 0$ for all $n>N$ .

and I'm kind of impressed with myself for learning all the latex to write this out

Ok so I'm thinking that without loss of generality I can let $z_o$ be 0
then I'm pretty sure I'd have to use the definition of uniform convergences to create a contradiction
but I don't know what to do for this
my first guess was to say that suppose $a_{N+1} \neq 0$
then do something with the inequality
$\mid S_N(x)-S(x) \mid < \varepsilon$
by using $S_{N+1}(x) = S_N(x) + a_{N+1}$

I think I might be close, but I don't know what to do from here

**choovuck** · September 28th 2008, 09:14 PM

assume there exist $\epsilon>0$ and number N such that $|S(x)-S_N(x)|<\epsilon$ for all $z\in\mathbb{C}$ . This means that $S(x)-S_N(x)=\sum_{n=N+1}^\infty a_n(z-z_0)^n$ is analytic and bounded on $\mathbb{C}$ , thus constant by Liouville. Thus $a_n=0$ for all $n\ge N+1$ .

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