Suppose the complex power series \(\textstyle \sum _ { n = 0 } ^ { \infty } a _ { n } z ^ { n }\) converges for some \(z _ { 0 } \neq 0\)
Prove that the series converges absolutely for each \(z\) with \(\abs{z} < \abs{z_0}\)
Suppose \(0 < r < | z _ { 0 } |\) . Show that the series converges uniformly on \(| z | \leq r\)
Solution
Since \[ \sum_{n=0}^{\infty}a_nz_0^n \] converges, its terms form a bounded sequence. Choose \(M\geq0\) such that \[ \abs{a_nz_0^n}\leq M \] for every \(n\geq0\).
1. If \(\abs{z}<\abs{z_0}\), then\[ \sum_{n=0}^{\infty}a_nz^n \]converges absolutely.
Set \[ q=\frac{\abs{z}}{\abs{z_0}}. \] Then \(0\leq q<1\), and for every \(n\), \[ \begin{aligned} \abs{a_nz^n} &= \abs{a_nz_0^n} \left(\frac{\abs{z}}{\abs{z_0}}\right)^n\\ &\leq Mq^n. \end{aligned} \] The geometric series \[ \sum_{n=0}^{\infty}Mq^n \] converges, so comparison proves \[ \sum_{n=0}^{\infty}\abs{a_nz^n}<\infty. \]
2. If \(0<r<\abs{z_0}\), then the series converges uniformly on\[ \abs{z}\leq r. \]
Put \[ q_r=\frac{r}{\abs{z_0}}, \] so \(0<q_r<1\). For every \(z\) with \(\abs{z}\leq r\) and every \(n\), \[ \begin{aligned} \abs{a_nz^n} &= \abs{a_nz_0^n} \left(\frac{\abs{z}}{\abs{z_0}}\right)^n\\ &\leq Mq_r^n. \end{aligned} \] The numerical series \[ \sum_{n=0}^{\infty}Mq_r^n \] converges. Hence the Weierstrass M-test gives uniform convergence on \(\abs{z}\leq r\).
3. Q.E.D.
Step 1 proves part (a), and step 2 proves part (b).