Complex Analysis/Identity Theorem

Proof Type

The equivalence is proven using a cyclic implication: ${\displaystyle (1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)}$

Proof (N1 to N2)

(N1) ${\displaystyle \Rightarrow }$ (N2) is obvious, as all derivatives of the zero function ${\displaystyle f}$ are zero.

Proof (N2 to N3)

Assume (N2). Consider the power series expansion ${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}}$ in ${\displaystyle B_{r}(z_{0})}$ with ${\displaystyle r>0}$. Here, ${\displaystyle a_{n}={\frac {f^{(n)}(z_{0})}{n!}}=0}$ for all ${\displaystyle n\in \mathbb {N} }$. Thus, ${\displaystyle f|_{B_{r}(z_{0})}=0}$, and (N3) follows.

Proof (N3 to N1) – Contradiction Proof

The step (N3) ${\displaystyle \Rightarrow }$ (N1) is proven by contradiction. Assume the zero set has a limit point and ${\displaystyle f}$ is not the zero function.

Proof 1 - (N3 to N1) - Power Series Expansion at Limit Point

Assume (N3), i.e., the set ${\displaystyle N_{f}}$ of zeros of ${\displaystyle f}$ has a limit point ${\displaystyle z_{0}\in U}$. Thus, there exists a sequence ${\displaystyle (z_{n})\in U^{\mathbb {N} }}$ with ${\displaystyle z_{n}\to z_{0}}$ and ${\displaystyle f(z_{n})=0}$ as well as ${\displaystyle z_{n}\neq z_{0}}$ for all ${\displaystyle n\in \mathbb {N} }$. Let ${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}}$ be the power series expansion of ${\displaystyle f}$ around ${\displaystyle z_{0}}$.

Proof 2 - (N3 to N1) - Power Series Expansion

Suppose there exists ${\displaystyle n\in \mathbb {N} }$ with ${\displaystyle a_{n}\neq 0}$. Due to the well-ordering property of ${\displaystyle \mathbb {N} }$, there would also be a smallest such ${\displaystyle n}$. Then

${\displaystyle f(z)=(z-z_{0})^{n}\sum _{k=0}^{\infty }a_{n+k}(z-z_{0})^{k},\qquad |z-z_{0}|<r,a_{n}\neq 0}$

Proof 3 - (N3 to N1) - Power Series Evaluation

For each ${\displaystyle i\in \mathbb {N} }$, we have

${\displaystyle 0=f(z_{i})=(z_{i}-z_{0})^{n}\sum _{k=0}^{\infty }a_{n+k}(z_{i}-z_{0})^{k}}$

Proof 4 - (N3 to N1) - Limit Process

Since ${\displaystyle z_{i}\neq z_{0}}$ and ${\displaystyle \displaystyle \lim _{i\to \infty }z_{i}=z_{0}}$, we get

${\displaystyle 0=\sum _{k=0}^{\infty }a_{n+k}(z_{i}-z_{0})^{k}\to a_{n},\qquad i\to \infty }$

As ${\displaystyle a_{n+k}(z_{i}-z_{0})^{k}\to 0}$ for all ${\displaystyle k>0}$ as ${\displaystyle i\to \infty }$. This contradicts ${\displaystyle a_{n}\neq 0}$. Therefore, ${\displaystyle a_{n}=0}$ for all ${\displaystyle n\in \mathbb {N} }$, and hence ${\displaystyle f^{(n)}(z_{0})=0}$ for all ${\displaystyle n\in \mathbb {N} }$, i.e., (N2) holds.

Proof 5 - (N3 to N1) - V is Closed

If (N2) holds, set ${\displaystyle V:=\bigcap _{n\geq 0}{z\in U,|,f^{(n)}(z)=0}}$. ${\displaystyle V}$ is closed in ${\displaystyle U}$ as the intersection of closed sets, because the ${\displaystyle f^{(n)}}$ are continuous, and preimages of closed sets (here ${\displaystyle {0}}$) are closed.

Proof 6 - (N3 to N1) - V is Open

${\displaystyle V}$ is also open in ${\displaystyle U}$, as for every ${\displaystyle w\in V}$, the power series expansion of ${\displaystyle f}$ around ${\displaystyle w}$ vanishes. Thus, ${\displaystyle f}$ is locally zero around ${\displaystyle w}$. Since ${\displaystyle z_{0}\in V}$, ${\displaystyle V}$ is non-empty, and hence ${\displaystyle V=U}$ due to the connectedness of ${\displaystyle U}$.

Proof 7 - From (N1)-(N3) to (1)-(3)

The statement of the Identity Theorem (1)-(3) follows for arbitrary ${\displaystyle g:U\to \mathbb {C} }$ and ${\displaystyle h:U\to \mathbb {C} }$, by applying (N1)-(N3) to ${\displaystyle f:=g-h}$.

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Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Kurs:Funktionentheorie/Identitätssatz - URL: https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Identitätssatz
• Date: 12/17/2024