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Math 416 Lecture 14

Review

Holomorphic \iff Analytic

Theorem 7.11 Liouville's Theorem

Any bounded entire function is constant.

New Rollings

Finding power series for holomorphic functions

Let F be holomorphic on open set U\subset \mathbb{C}. Suppose f(z_0)=0, f(z)=\sum_{n=0}^\infty a_n(z - z_0)^n

Example,

p(z)=(z-1)^3(z+i)^5(z-7)

p(z)=\sum_{n=0}^9 c_n(z-z_0)^n

Notice that:

Since f'(z)=\sum_{n=0}^\infty a_n n(z-z_0)^{n-1}. a_0=f(z_0), a_1=f'(z_0), a_k=\frac{f^{(k)}(z_0)}{k!} for k \geq 0

So c_0=0=f(1), c_1=f'(1)=3(z-1)^2M=0, c_2=f''(1)=6(z-1)M=0, c_3=1.

(i) The power series for q(z)=(z-1)^3 at 0.

So q(z)=\sum_{n=0}^3 a_nz^n, you just expand it as q(z)=z^3-3z^2+3z-1

(ii) The power series for q(z)=(z-1)^3 at -1.

So q(z)=\sum_{n=0}^3 a_n(z+1)^n

a_0=q(-1)=(-2)^3=-8,

a_1=q'(-1)=3(-2)^2=12,

a_2=\frac{1}{2}q''(-1)=\frac{6}{2} \cdot -2^1 = -6,

a_3=\frac{1}{6}q'''(-1)=\frac{6}{6} \cdot 1=1.

All higher terms are zero

Definition: zero of multiplicity

Suppose f is holomorphic on open U and f(\zeta_0)=0 for some z_0\in U. Let f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n near z_0. Let m be the smallest number such that a_m\neq 0. Then we say f has a zero of multiplicity m at z_0.

Theorem 7.12 Fundamental Theorem of Algebra

Every non-constant polynomial f can be factored over \mathbb{C} into linear factors

Proof:

Since a_n=\frac{1}{n!}f^{(n)}(z_0), then f has a zero of order m \iff f^{(m)}(z_0) \neq 0 and f^{(k)}(z_0) = 0, \forall k < m.

Suppose f has a zero of order m at z_0

\begin{aligned}
f(z)&=a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots\\
&=(z-z_0)^m\left[a_{m+1}(z-z_0)^{m+1}+\cdots\right]\\
&= (z-z_0)^m g(z)
\end{aligned}

So, if f has a zero of order m at \zeta_0\iff f(z)=(z-z_0)^mg(z) where g is holomorphic and g(z_0)\neq 0.

QED

Definition: Connected Set

An open set U is connected if whenever U=U_1\cup U_2 and U_1, U_2 are disjoint and open, then one of them is empty.

A domain is a connected open set.

Theorem 7.13 Zeros of Holomorphic Functions

Let U be a open domain (in \mathbb{C}). Let f be holomorphic on U and vanish to infinite order at some point z_0\in U, then f(z)=0 on U.

This is not true for \mathbb{R}. Consider the function f(x) = e^{-1/x^2} for x \neq 0 and f(0) = 0, which is smooth and vanishes to infinite order at 0.

Proof:

Step 1:

Show any zero of finite order is isolated.

Let z_0 be a zero of order m, then by fundamental theorem of algebra, f can be expressed as

f(z)=(z-z_0)^mg(z)

where g is holomorphic and g(z_0) \neq 0. So g is continuous.

Thus \exists and open set z_0\in V such that g(z_0)\neq 0 on all of V.

Let U_1=\{z\in U\} such that f vanishes to order infinity. and U_2=U\setminus U_1.

We need to show both U_1 and U_2 are open.

U_1:

Let z_0\in U_1. We know that f is holomorphic thus it is analytic at z_0.

So \exists \epsilon>0 such that \forall \mathbb{z}\in B_\epsilon(z_0)

So f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n implies f(z)=0 on B_\epsilon(z_0)

We can expand f in a power series centered at z_1 for any z_1\in B_\epsilon(z_0), So f(z)=\sum c_n(z-z_1)^n=0

Therefore, z_1 \in U_1, proving that U_1 is open.

U_2:

Let w\in U_2, if f(w)\neq 0, then \exists\epsilon > 0 such that f(z) \neq 0 on B_\epsilon(w)\subset U_2.

If f vanishes to finite order by Step 1, \exists B_\epsilon(w)\subset U_2

QED

Corollary 7.13.1 (Identity for holomorphic functions)

If f,g are both holomorphic on domain U, and they have the same power series at some point \zeta_0, then f \equiv g on U.

Proof:

Consider f-g.

QED

Corollary 7.13.2

Let U be a domain, f\in O(U), f is not identically zero on U, f^{-1}(0) has no limit point on U.

Proof:

We proceed by contradiction. Suppose z_n\to w\in U, f(z_0)=0, f(w)=0. w is not an isolated zero. So f is a zero of infinite order. Contradicting with our assumption that f is not identically zero.

QED

Corollary 7.14: Identity principle

If f,g\in O(U), U is a domain and \exists sequence z_0 that converges to w\in U, such that f(z_n)=g(z_n), then f\equiv g on U$.