proof of identity theorem of power series
We start by proving a more modest result. Namely, we show that, under the hypotheses of the theorem we are trying to prove, we can conclude that ${a}_{0}={b}_{0}$.
Let $R$ be chosen such that both series converge^{} when $$. From the set of points at which the two power series^{} are equal, we may choose a sequence ${\{{w}_{k}\}}_{k=0}^{\mathrm{\infty}}$ such that

•
$$ for all $k$.

•
${lim}_{k\to \mathrm{\infty}}{w}_{k}$ exists and equals ${z}_{0}$.

•
${w}_{k}\ne {z}_{0}$ for all $k$.
.
Since power series converge uniformly, we may interchange the limit with the summation.
$\underset{k\to \mathrm{\infty}}{lim}{\displaystyle \sum _{n=0}^{\mathrm{\infty}}}{a}_{n}{({w}_{k}{z}_{0})}^{n}$  $=$  $\sum _{n=0}^{\mathrm{\infty}}}\underset{k\to \mathrm{\infty}}{lim}{a}_{n}{({w}_{k}{z}_{0})}^{n}={a}_{0$  
$\underset{k\to \mathrm{\infty}}{lim}{\displaystyle \sum _{n=0}^{\mathrm{\infty}}}{b}_{n}{({w}_{k}{z}_{0})}^{n}$  $=$  $\sum _{n=0}^{\mathrm{\infty}}}\underset{k\to \mathrm{\infty}}{lim}{b}_{n}{({w}_{k}{z}_{0})}^{n}={b}_{0$ 
Because ${\sum}_{n=0}^{\mathrm{\infty}}{a}_{n}{({w}_{k}{z}_{0})}^{n}=su{m}_{n=0}^{\mathrm{\infty}}{a}_{n}{({w}_{k}{z}_{0})}^{n}$ for all $k$, this means that ${a}_{0}={b}_{0}$.
We will now prove that ${a}_{n}={b}_{n}$ for all $n$ by an induction^{} argument^{}. The intial step with $n=0$ is, of course, the result demonstrated above. Assume that ${a}_{m}={b}_{m}$ for all $m$ less than some integer $N$. Then we have
$$\sum _{n=N}^{\mathrm{\infty}}{a}_{n}{(w{z}_{0})}^{n}=\sum _{n=N}^{\mathrm{\infty}}{b}_{n}{(w{z}_{0})}^{n}$$ 
for all $w\in S$. Pulling out a common factor and relabelling the index, we have
$${(w{z}_{0})}^{N}\sum _{n=0}^{\mathrm{\infty}}{a}_{n+N}{(w{z}_{0})}^{n}={(w{z}_{0})}^{N}\sum _{n=0}^{\mathrm{\infty}}{b}_{n+N}{(w{z}_{0})}^{n}.$$ 
Because ${z}_{0}\notin S$, the factor $w{z}_{0}$ will not equal zero, so we may cancel it:
$$\sum _{n=0}^{\mathrm{\infty}}{a}_{n+N}{(w{z}_{0})}^{n}=\sum _{n=0}^{\mathrm{\infty}}{b}_{n+N}{(w{z}_{0})}^{n}$$ 
By our weaker result, we have ${a}_{N}={b}_{N}$. Hence, by induction, we have ${a}_{n}={b}_{n}$ for all $n$.
Title  proof of identity theorem of power series^{} 

Canonical name  ProofOfIdentityTheoremOfPowerSeries 
Date of creation  20130322 16:47:38 
Last modified on  20130322 16:47:38 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  7 
Author  rspuzio (6075) 
Entry type  Proof 
Classification  msc 30B10 
Classification  msc 40A30 