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 $z_0$

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

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$.