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.
