Complex analysis review 6

Laurent Series

The special properties of a complex function is much more determined by its singularity, to study the singularity of a function, we first give a useful theorem that does not hold for real functions.

Theorem 1 (Weierstrass)

Suppose that $\{f_j(z)\}$ are analytic on $U\subset \mathbb{C}$, and $\sum_{n=1}^\infty f_n(z)$ uniformly converge on any closed subset of $U$ to $f(z)$, then $f(z)$ is analytic on $U$, and $\sum_{n=1}^\infty f_n^{(k)}(z)$ converges uniformly to $f^{(k)}(z)$ on any closed subset of $U$.

For any closed rectifiable simple curve $\gamma$ on $U$, we have

$$\int_\gamma f(z) dz = \sum_{n=1}^\infty \int_\gamma f_n(z)dz=0.$$
Then by Morera theorem, $f(z)$ is analytic on $U$.

If $z_0\in U, \bar{D}(z_0,r)\subset U$, then $\sum_{n=1}^\infty f_n(z)$ converges to $f(z)$ uniformly on $\partial D(z_0,r)$. So for $z\in D(z_0, r/2)$,

$$\sup_{z\in \bar{D}(z_0, r/2)}|\sum_{j=1}^nf_j^{(k)}(z)-f^{(k)}(z)|\leq c_n \sup_{z\in D(z_0,r)}|\sum_{j-1}^nf_j(z)-f(z)|.$$
Then we can get a open covering and use Heine-Borel theorem, the conclusion is true for any closed bounded subset of $U$.

Now we can define a Laurent series at $a\in \mathbb{C}$,

$$\sum_{n=-\infty}^\infty c_n (z-a)^n.$$
$\sum_{n=0}^\infty c_n(z-a)^n$ is called the analytic part of the above series, and the remain is called the principle part.

Theorem 2

If $f(z)$ is analytic on $V: r<|z-a|<R, (0\leq r <R<\infty)$, then $f$ has a unique Laurent expansion on $V$

$$f(z) = \sum_{n=-\infty}^\infty c_n(z-a)^n,$$
where
$$c_n = \frac{1}{2\pi i}\int_{|\xi-a|=\rho}\frac{f(\xi)}{(\xi-a)^{n+1}}d\xi \quad (r<\rho <R).$$

Isolated Singular Point

If $f$ is analytic on a neighbourhood $D(a,R)\backslash \{a\}$ of $a$, then $a$ is called a isolated singular point of $f$.

From theorem 2, there is a Laurent series of $f$ on $0<|z-a|< R$,

$$f(z) = \sum_{n=-\infty}^\infty c_n(z-a)^n.$$
There are three case to be considered.

Removable singular point

$\lim_{z\rightarrow a} f(z)$ exists and finite, from Riemann theorem, $f$ can be extended to an analytic functon on $D(a,R)$, so $c_{-n}$ are all zeros.

Poles of order $m$

$\lim_{z\rightarrow a} f(z)$ exists but infinite, then only finitely number of $c_{-n}$ are nonzero. Therefore

$$f(z) = \frac{c_{-m}}{(z-a)^m}+\cdots+\frac{c_{-1}}{z-a}+\sum_{n=0}^\infty c_n(z-a)^n.$$
Since $\lim_{z\rightarrow a}f(z)=\infty$, so there is $\delta>0$. such that $f(z)\neq 0, 0<|z-a|<\delta$, so on this field, $F(z) = \frac{1}{f(z)}$ is analytic and nonzero, moreover $\lim_{z\rightarrow a} F(z)=0$. Therefore $a$ is removable singular point of $F(z)$, then
$$F(z) = (z-a)^m\lambda (z).$$
Without lost of generality, we assume that $1/\lambda(z)$ is nonzero on $|z-a|<\delta$. Then it has Taylor expansion
$$\frac{1}{\lambda(z)} = c_{-m}+c_{-m+1}(z-a)+\cdots$$

Essential singular point

$\lim_{z\rightarrow a} f(z)$ does not exist.

In this case, $c_{-n}$ have infinite terms which are nonzero.

Theorem 3 (Weierstrass)

If $a$ is an essential singular point of $f$, for any given $\delta>0$, and any complex number $A$, $\epsilon>0$, there is a $z$ on $0<|z-a|<\delta$, such that

$$|f(z)-A|<\epsilon.$$
Which means that the values of $f$ near essential singular point is dense in $\mathbb{C}$.

This can be showed easily by prove the converse.

Residual Theorem

Define residual of $f$, which is analytic on $D(z,r)\backslash \{a\}$,

$$Res(f,a) = \frac{1}{2\pi i}\int_{|z-a|=\rho} f(z)dz\quad (0<\rho<r).$$
Use Laurent series we can deduce that
$$Res(f,a) = c_{-1}.$$
If $z=\infty$ is isolated singular point of $f$, and $f$ is analytic on $R<|z|<\infty$, then
$$Res(f,\infty)=-c_{-1}.$$
If $a(a\neq \infty)$ is pole of $f$ of order $m$, then
$$f(z) =\frac{1}{(z-a)^m}g(z),$$
where $g(z)$ is analytic on $a$, and $g(a)\neq 0$, then
$$g(z) = \sum_{n=0}^\infty \frac{1}{n!}g^{(n)}(a)(z-a)^n.$$
So
$$Res(f,a) = c_{-1} = \frac{1}{(m-1)!}g^{(m-1)}(a)= \frac{1}{(m-1)!}\lim_{z\rightarrow a}\frac{d^{(m-1)}}{dz^{(m-1)}}[(z-a)^mf(z)].$$

Theorem 4

If $f$ is analytic on $\mathbb{C}\supset U\backslash \{z_1,\cdots,z_n\}$, and continuous on $\mathbb{C}\supset \bar{U}\backslash \{z_1,\cdots,z_n\}$, $\partial U$ is a simple closed rectifiable curve, then

$$\int_{\partial U}f(z)dz = 2\pi i \sum_{k=1}^n Res(f,z_k).$$