Complex Analysis Ⅲ

1.Power Series

Definition 1.1
In general, a power series is the expansion of $sum_{n=0}^{infty}a_nz^n$ where .

Next, we will give a theorem on the convergence of the power series.

Theorem 1.2 (Cauchy-Hadmard Theorem)
Given a power series $sum_{n=0}^{infty}a_nz^n$ , there exists a such that (1) when , the series converges absolutely; (2) when , the series diverges.

If we use the convention , $R^{-1}=limsup |a_n|^{1/n}$ .

Proof: When , $lim_{n oinfty}|a_n|^{1/n}=infty$ . So for all , there exists infinitely many $|a_n|^{1/n}>|z|^{-1}$ . So which means the series diverges.

When , $lim_{n oinfty}|a_n|^{1/n}=0$ . So for all $0<varepsilon<|z|^{-1}$ , $|a_n|^{1/n}<varepsilon$ when . Hence . Since , the series converges absolutely.

Otherwise, when , choose a small such that $(R^{-1}+varepsilon)|z|=r<1$ . Since $limsup |a_n|^{1/n}=R^{-1}$ , there exists infinitely many $|a_n|^{1/n}<R^{-1}+varepsilon$ . So . Since converges, the series converges absolutely. When , for any , there exists $|a_n|>(R^{-1}-varepsilon)^n$ . If $|z|=frac{1}{R^{-1}-c}>R$ , and hence the series diverges.

Q.E.D.

2.Analytic Function

Theorem 2.1
Let $f(z)=sum_{n=0}^{infty}a_nz^n$ and $g(z)=sum_{n=0}^{infty}na_nz^{n-1}$ . and have the same convergence disc. is holomorphic on its convergence disc and .

Proof: Because $limsup |a_n|^{1/n} ot=infty$ , $limsup |na_n|^{1/n}=limsup |a_n|^{1/n}$ since $lim_{n oinfty}n^{1/n}=1$ . This shows that and have the same convergence disc.

Let where $S_N(z)=sum_{n=1}^Na_nz^n$ and $E_N(z)=sum_{n=N+1}^{infty}a_nz^n$ . Hence, $left|frac{f(z+h)-f(z)}{h}-g(z) ight|\ leqleft|frac{S_N(z+h)-S_N(z)}{h}-S_N(z) ight|+left|frac{E_N(z+h)-E_N(z)}{h} ight|+|S_N(z)-g(z)|$

Obviously, there exists a and a such that $left|frac{S_N(z+h)-S_N(z)}{h}-S_N(z) ight|<frac{varepsilon}{3}$ and $|S_N(z)-g(z)|<frac{varepsilon}{3}$ when .

Next, $egin{equation} left|frac{E_N(z+h)-E_N(z)}{h} ight|=left|frac{sum_{n=N+1}^{infty}a_n[(z+h)^n-z^n]}{h} ight|=left|sum_{n=N+1}^{infty}a_nsum_{i=0}^{n-1}(z+h)^iz^{n-i-1} ight| end{equation} (1)$

Assume that and . So (1) is less than $left|sum_{n=N+1}^{infty}na_nr^{n-1} ight|$ . Since converges when , there exists a such that (1) is less than $frac{varepsilon}{3}$ . Therefore, $left|frac{f(z+h)-f(z)}{h}-g(z) ight|<varepsilon$ . So .

Q.E.D.

Corollary 2.2
A power series is infinitely differentiable on its convergence disc.

Definition 2.3
A function is analytic at if there is a power series $sum_{n=0}^{infty}a_n(z-z_0)^n$ centered at with positive radius of convergence and $f(z)=sum_{n=0}^{infty}a_n(z-z_0)^n$ for all in a neighbourhood of . If is analytic at all , then is analytic on .