Let \(f(z) = \sum_{n=0}^\infty a_n z^n\) be a power series centered at the origin with radius of convergence \(R > 0\). Prove that \(f\) has a power series expansion about any point \(z_0\) in its disc of convergence \(D_R(0)\).
Solution
Fix \(z_0 \in D_R(0)\) and \(r\) with \(0 < r < R - \abs{z_0}\). Choose \(\rho\) with \(\abs{z_0} + r < \rho < R\), and let \(\Gamma\) be the circle \(\abs{\xi} = \rho\), oriented counterclockwise.
1. The function \(f\) is holomorphic on \(D_R(0)\), and every \(z \in D_r(z_0)\) lies inside \(\Gamma\).
A power series converges uniformly on compact subsets of its open disc of convergence, and its sum is holomorphic there. For \(z \in D_r(z_0)\), the triangle inequality gives \(\abs{z} \le \abs{z_0} + \abs{z - z_0} < \abs{z_0} + r < \rho\).
2. For every \(z \in D_r(z_0)\),\[f(z) = \frac{1}{2\pi i} \oint_\Gamma \frac{f(\xi)}{\xi - z} \, d\xi.\]
By step 1, \(f\) is holomorphic on \(D_R(0)\supset\overline{D}_\rho(0)\) and \(z\) lies inside \(\Gamma\), so Cauchy’s integral formula applies.
3. For every \(z \in D_r(z_0)\), the series\[\frac{f(\xi)}{\xi - z} = \sum_{k=0}^\infty \frac{f(\xi)}{(\xi - z_0)^{k+1}} (z - z_0)^k\]converges uniformly for \(\xi \in \Gamma\).
For \(\xi\in\Gamma\), \[\frac{1}{\xi - z} = \frac{1}{\xi - z_0} \cdot \frac{1}{1 - \frac{z - z_0}{\xi - z_0}} \qquad\text{and}\qquad \abs{\xi - z_0} \ge \abs{\xi} - \abs{z_0} = \rho - \abs{z_0}.\] Hence \[\left| \frac{z - z_0}{\xi - z_0} \right| \le c \coloneqq \frac{r}{\rho - \abs{z_0}} < 1,\] and the Weierstrass \(M\)-test with \(M_k=c^k\) shows that the geometric series \(\sum_k \bigl(\frac{z - z_0}{\xi - z_0}\bigr)^k\) converges uniformly on \(\Gamma\). Multiplying by \(f(\xi)/(\xi - z_0)\), which is bounded on \(\Gamma\), preserves uniform convergence.
4. For every \(z \in D_r(z_0)\),\[f(z) = \sum_{k=0}^\infty b_k (z - z_0)^k, \qquad b_k = \frac{1}{2\pi i} \oint_\Gamma \frac{f(\xi)}{(\xi - z_0)^{k+1}} \, d\xi = \frac{f^{(k)}(z_0)}{k!}.\]
Substitute step 3 into step 2 and integrate term by term, which uniform convergence on \(\Gamma\) permits. The second formula for \(b_k\) is Cauchy’s integral formula for derivatives.
5. Q.E.D.
The coefficients \(b_k = f^{(k)}(z_0)/k!\) do not depend on \(r\). Since \(r < R - \abs{z_0}\) was arbitrary, step 4 gives the expansion \(f(z) = \sum_{k} b_k (z - z_0)^k\) on the whole disc \(D_{R - \abs{z_0}}(z_0)\).