Complex analysis review 4

The zeros of analytic functions

Use Liouville theorem, we can prove the fundamental theorem of algebra.

Theorem 1

If p(z) is a polynomial, then there is at least one z0 such that p(z0)=0 .

Let

f(z)=1p(z)

then if p has no roots, f is analytic on C . Implying that f is a constant, so as p .

Theorem 2

Suppose that f(z) is analytic on U⊂C , then except for being the constant function 0, there is no accumulation point of the set {z∈U|f(z)=0} .

If it has a accumulation point, i.e., suppose that z1,⋯,zn,⋯ , are zeros of f , and there is an accumulation point z0 . Without loss of generality, we assume that z0=0 . Then since f is analytic on U , there is a Taylor expansion of f at point 0

f(z)=a0+a1z+⋯

Then

limn→∞f(zn)=f(limn→∞zn)=f(0)=a0=0.

Next we can consider f(z)/z , for the same reason we have a1=0 . The process continues.

Now we can conclude that if E is a subset of U and E has an accumulation point, h1=h2 on E . Then h1=h2 on U . So for some triangular equations, if they hold in R , then they also hold in C .

The argument principle

Theorem 3

If f(z) is an analytic function inside and on some closed contour γ , and f has no zeros on γ as well as finitely many zeros inside the contour. Then the number k of zeros insider the contour γ is

k=12πi∫γf′(z)f(z)dz.

Denote the zeros and their multiple number as z1,⋯,zn , with respect to k1,⋯,kn . They by Cauchy integral formula,

12πi∫γf′(z)f(z)dz=∑i=1n12πi∫γif′(z)f(z)dz.

Inside of each γi , we can write

f(z)=(z−zi)kihi(z).

Where hi is nonzero, and then

f′(z)f(z)=kiz−zi+h′i(z)hi(z)

which give the desired result.

Theorem 4 (Hurwitz)

Suppose that {fj} is a sequence of analytic function on U⊂C , for any compact subset of U , it converges uniformly to a function f . If all of the function in the sequence are not identity to zero, then f is either equals to constant function 0 or never equals to zero.

Choose any closed contour γ , then for z insider the contour

fj(z)=12πi∫γfj(ξ)ξ−zdξ

By hypothesis, let j→∞ , then

f(z)=12πi∫γf(ξ)ξ−zdξ

So f(z) is an analytic function and by the same reason f′j converges to f′ on any compact subset of U . If f is not constant function 0 then we choose γ not pass through its zeros, which is guaranteed by the discreteness of zeros of analytic functions. And note that

12πi∫γf′j(z)fj(z)dz=0.

We concluded that

12πi∫γf′(z)f(z)dz=0.

Rouche’s Theorem

Theorem 4

Suppose that f(z),g(z) are analytic on U⊂C , γ is a contour that is rectifable, on γ there holds

|f(z)−g(z)|<|f(z)|,

then f,g have the same number zeros insider the contour γ .

Let N1,N2 denote the number of zeros of f,g respectively. Then

N2−N1=12πi∫γF′(z)F(z)dz.

Where F(z)=g(z)f(z) . And from the hypothese, we know that

|F(z)−1|<1.

Which shows that N2−N1=0 .

Let

p(z)=anzn+⋯+a0,an≠0.

Choose g(z)=anzn , then g has n roots. When R is large enough, on contour |z|=R , we have

|p(z)−g(z)|<|g(z)|=|an|Rn.

By Rouche’s theorem, p has exactly n roots.

Theorem 5

Suppose that f is analytic on U⊂C , w0=f(z0),z0∈U . If z0 has multiple number m , then for ρ>0 small enough, there is δ>0 , such that for any A∈D(w0,δ) , the number of zeros of f(z)−A on D(z0,ρ) is exactly m .

There is some ρ>0 small enough such that f(z)−f(z0) has no zeros except for z0 on D¯(z0,ρ)⊂U . But in the circle |z−z0|=ρ , there holds

|f(z)−f(z0)|≥δ(δ>0).

Then for A∈D(w0,δ) ,

|A−w0|<|f(z)−f(z0)|.

That is when |z−z0|=ρ ,

|(f(z)−f(z0))−(f(z)−A)|<|f(z)−f(z0)|.

This shows that f(z)−A has the same number of zeros in D(z0,ρ) which is m .