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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 $zz_{0}<R$. From the set of points at which the two power series are equal, we may choose a sequence $\{w_{k}\}_{{k=0}}^{\infty}$ such that

$w_{k}z_{0}<R/2$ for all $k$.

$\lim_{{k\to\infty}}w_{k}$ exists and equals $z_{0}$.

$w_{k}\neq z_{0}$ for all $k$.
.
Since power series converge uniformly, we may interchange the limit with the summation.
$\displaystyle\lim_{{k\to\infty}}\sum_{{n=0}}^{\infty}a_{n}(w_{k}z_{0})^{n}$  $\displaystyle=$  $\displaystyle\sum_{{n=0}}^{\infty}\lim_{{k\to\infty}}a_{n}(w_{k}z_{0})^{n}=a_% {0}$  
$\displaystyle\lim_{{k\to\infty}}\sum_{{n=0}}^{\infty}b_{n}(w_{k}z_{0})^{n}$  $\displaystyle=$  $\displaystyle\sum_{{n=0}}^{\infty}\lim_{{k\to\infty}}b_{n}(w_{k}z_{0})^{n}=b_% {0}$ 
Because $\sum_{{n=0}}^{\infty}a_{n}(w_{k}z_{0})^{n}=sum_{{n=0}}^{\infty}a_{n}(w_{k}z_% {0})^{n}$ 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
$\sum_{{n=N}}^{\infty}a_{n}(wz_{0})^{n}=\sum_{{n=N}}^{\infty}b_{n}(wz_{0})^{n}$ 
for all $w\in S$. Pulling out a common factor and relabelling the index, we have
$(wz_{0})^{N}\sum_{{n=0}}^{\infty}a_{{n+N}}(wz_{0})^{n}=(wz_{0})^{N}\sum_{{n% =0}}^{\infty}b_{{n+N}}(wz_{0})^{n}.$ 
Because $z_{0}\notin S$, the factor $wz_{0}$ will not equal zero, so we may cancel it:
$\sum_{{n=0}}^{\infty}a_{{n+N}}(wz_{0})^{n}=\sum_{{n=0}}^{\infty}b_{{n+N}}(wz% _{0})^{n}$ 
By our weaker result, we have $a_{N}=b_{N}$. Hence, by induction, we have $a_{n}=b_{n}$ for all $n$.
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