Complex analysis review 3

Taylor series and Liouville theorem

Many interesting results can be obtained by the Cauchy Integral Theory.

Theorem 1

Suppose that $f(z)$ is analytic on $U\subset C$, and continuous on $\bar{U}$. Then at any point in $U$, and any non-negative integer $n$, the $n-$ order derivative exists and

$$f^{(n)}(z) = \frac{n !}{2\pi i}\int_{\partial U}\frac{f(\xi)}{(\xi-z)^{n+1}}d\xi.$$
Moreover, if $z_0\in U,\bar{D}(z_0,r) \subset U$, then $f$ has a Taylor expansion on $D(z_0,r)$
$$f(z) = \sum_{j=0}^\infty a_j(z-z_0)^j.$$
And the Taylor series is absolutely and uniformly convergece on $\bar{D}(z_0,r)$. Furthermore,
$$a_j = \frac{1}{2\pi i}\int_{\partial U}\frac{f(\xi)}{(\xi -z_0)^{j+1}}d\xi.$$
Proof:

When $z\in D(z_0,r)$, by Cauchy integral theorem

$$\frac{f(z)-f(z_0)}{z-z_0}-\frac{1}{2\pi i}\int_{\partial U}\frac{f(\xi)}{(\xi-z)^2}d\xi = \frac{z-z_0}{2\pi i}\int_{\partial U}\frac{f(\xi)}{(\xi-z){(\xi - z_0)}}d\xi$$
Let
$$d = \inf_{z\in \partial U}||z-z_0||_2.$$
Then for $r = d/2$,
$$|\xi-z|\geq d/2\\ |\xi-z_0|\geq d$$
So
$$|\int_{\partial U}\frac{f(\xi)}{(\xi-z){(\xi - z_0)}}d\xi|\leq \frac{2ML}{d^3}$$
Where $M = \max_{\xi\in \partial U}|f(\xi)|$, $L$ is the length of $\partial U$.

The following part of proof is done by induction.

Theorem 2 (Cauchy Inequality)

Suppose that $f(z)$ is analytic on $U\subset C$, $\bar{D}(z_0,R)\subset U$. Then

$$|f^{(j)}(z)|\leq \frac{j!M}{R^j}.$$
Where $M = \max_{z\in \bar{D}(z_0,R)}|f(z)|$.

The result is directly from the previous theorem.

Theorem 3

Suppose that $U\subset C$, $K$ is a compact set in $U$, $V\supset K$ is relatively compact in $U$, then for any analytic function $f(z)$ on $U$, there are constants $c_n$ , such that

$$\sup_{z\in K}|f^{(n)}(z)|\leq c_n||f||_{L(V)}\\ ||f||_{L(V)} = \frac{1}{|V|}\iint_V |f(\xi)|dA$$
When $U$ is a compact set, then $f$ attains its maximum modulus on $U$. Therefore
$$\sup_{z\in K}|f^{(n)}(z)|\leq c_n\sup_{z\in V}|f(z)|,\quad \forall n.$$

Theorem 4

On the contrary, we have the following theorem for determing if a function is analytic.

(Morera Theorem) Suppose that $f$ is continuous on $U$, and its integral along any rectifiable simple closed curve is 0, then $f$ is analytic on $U$.

Proof:

Choose any point $z_0$, and define

$$F(z) = \int_{z_0}^zf(\xi)d\xi\quad (z\in U).$$
Then the integral is independent to the choice of path of integration. And
$$F'(z) = f(z)$$
So $F$ is analytic on $U$, so does $f$.

Theorem 5

Liouville theorem shows that if an analytic function on $C$ is also bounded, then it is a constant function.

(Liouville Theorem) If $f(z)$ is analytic on $C$ and bounded, then $f$ is a constant.

Since $|f(z)|\leq M$, we have $|f'(z)|=0$ .

Theorem 6

Suppose that $F$ is analytic and bounded on $D(z_0,r)\backslash \{z_0\}$, then there is a analytic function on $D(z_0,r)$ such that

$$f|_{D(z_0,r)\backslash \{z_0\}} =F.$$
Without loss of generality, suppose that $z_0=0$. Then define
$$G(z) =\left\{ \begin{align*} &z^2 F(z) &\quad if \, z\in D(z_0,r)\backslash \{z_0\}\\ &0\quad &if\, z=0 \end{align*} \right.$$
Then we can show that $G$ is analytic.

And let $f(z) = D(z)/z^2$.