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 {fj(z)} are analytic on U⊂C, and ∑∞n=1fn(z) uniformly converge on any closed subset of U to f(z), then f(z) is analytic on U, and ∑∞n=1f(k)n(z) converges uniformly to f(k)(z) on any closed subset of U.

For any closed rectifiable simple curve γ on U, we have

∫γf(z)dz=∑n=1∞∫γfn(z)dz=0.

Then by Morera theorem, f(z) is analytic on U.

If z0∈U,D¯(z0,r)⊂U, then ∑∞n=1fn(z) converges to f(z) uniformly on ∂D(z0,r). So for z∈D(z0,r/2),

supz∈D¯(z0,r/2)|∑j=1nf(k)j(z)−f(k)(z)|≤cnsupz∈D(z0,r)|∑j−1nfj(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∈C,

∑n=−∞∞cn(z−a)n.

∑∞n=0cn(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≤r<R<∞), then f has a unique Laurent expansion on V

f(z)=∑n=−∞∞cn(z−a)n,

where

cn=12πi∫|ξ−a|=ρf(ξ)(ξ−a)n+1dξ(r<ρ<R).

Isolated Singular Point

If f is analytic on a neighbourhood D(a,R)∖{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)=∑n=−∞∞cn(z−a)n.

There are three case to be considered.

Removable singular point

limz→af(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

limz→af(z) exists but infinite, then only finitely number of c−n are nonzero. Therefore

f(z)=c−m(z−a)m+⋯+c−1z−a+∑n=0∞cn(z−a)n.

Since limz→af(z)=∞, so there is δ>0. such that f(z)≠0,0<|z−a|<δ, so on this field, F(z)=1f(z) is analytic and nonzero, moreover limz→aF(z)=0. Therefore a is removable singular point of F(z), then

F(z)=(z−a)mλ(z).

Without lost of generality, we assume that 1/λ(z) is nonzero on |z−a|<δ. Then it has Taylor expansion

1λ(z)=c−m+c−m+1(z−a)+⋯

Essential singular point

limz→af(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 δ>0, and any complex number A, ϵ>0, there is a z on 0<|z−a|<δ, such that

|f(z)−A|<ϵ.

Which means that the values of f near essential singular point is dense in C.

This can be showed easily by prove the converse.

Residual Theorem

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

Res(f,a)=12πi∫|z−a|=ρf(z)dz(0<ρ<r).

Use Laurent series we can deduce that

Res(f,a)=c−1.

If z=∞ is isolated singular point of f, and f is analytic on R<|z|<∞, then

Res(f,∞)=−c−1.

If a(a≠∞) is pole of f of order m, then

f(z)=1(z−a)mg(z),

where g(z) is analytic on a, and g(a)≠0, then

g(z)=∑n=0∞1n!g(n)(a)(z−a)n.

So

Res(f,a)=c−1=1(m−1)!g(m−1)(a)=1(m−1)!limz→ad(m−1)dz(m−1)[(z−a)mf(z)].

Theorem 4

If f is analytic on C⊃U∖{z1,⋯,zn}, and continuous on C⊃U¯∖{z1,⋯,zn}, ∂U is a simple closed rectifiable curve, then

∫∂Uf(z)dz=2πi∑k=1nRes(f,zk).