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 $|z - z_0| < R$ . 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. \begin{eqnarray*} \lim_{k \to \infty} \sum_{n=0}^\infty a_n (w_k - z_0)^n &=& \sum_{n=0}^\infty \lim_{k \to \infty} a_n (w_k - z_0)^n = a_0 \\ \lim_{k \to \infty} \sum_{n=0}^\infty b_n (w_k - z_0)^n &=& \sum_{n=0}^\infty \lim_{k \to \infty} b_n (w_k - z_0)^n = b_0 \end{eqnarray*}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 $$ \sum_{n=N}^\infty a_n (w - z_0)^n = \sum_{n=N}^\infty b_n (w - z_0)^n $$ for all $w \in S$ . Pulling out a common factor and relabelling the index, we have $$ (w - z_0)^N \sum_{n=0}^\infty a_{n+N} (w - z_0)^n = (w - z_0)^N \sum_{n=0}^\infty b_{n+N} (w - z_0)^n. $$ Because $z_0 \notin S$ , the factor $w - z_0$ will not equal zero, so we may cancel it: $$ \sum_{n=0}^\infty a_{n+N} (w - z_0)^n = \sum_{n=0}^\infty b_{n+N} (w - z_0)^n $$ By our weaker result, we have $a_N = b_N$ . Hence, by induction, we have $a_n = b_n$ for all $n$ .