Notes - Complex Analysis MT23, Identity theorem

Created: November 19, 2023

Flashcards

Can you state the identity theorem for holomorphic functions?

Suppose:

  • $U$ is a domain
  • $f _ 1, f _ 2 : U \to \mathbb C$ are holomorphic functions
  • $S = \{z \in U \mid f _ 1(z) = f _ 2(z)\}$
  • $S$ has a limit point in $U$

Then

  • $f _ 1 = f _ 2$.

How could you very quickly justify that the set

\[A = \\{z \in U \mid f^{(n)}(z) = 0, \forall n \ge 0\\}\]

where $f$ is a holomorphic function is closed?

Note that $A _ n := \{z \in \mathbb U \mid f^{(n)}(z) = 0\}$ is the preimage of $\{0\}$ under $f^{(n)}$, which is continuous. Since $\{0\}$ is closed, $A _ n$ must also be closed.

Then

\[A = \bigcap_{n \ge 0} A_n\]

and so $A$ is also closed.

Suppose:

  • $U = B(0, 1)$
  • $f : U \to \mathbb C$ holomorphic
  • $f\left(\frac{n - 1}{n}\right) = 1 - \frac 2 n + \frac{1}{n^2}$

One such $f$ satisfying these properties is

\[f(z) = z^2\]

Can you give an alternative function $g$ with these properties and explain why the identity theorem does not apply here?
\[g(z) = z^2 + \sin\left(\frac{\pi}{1 - z}\right)\]

The constancy theorem does not apply since the points where $f(z) = g(z)$ has a limit point at $1$, but this is not in the domain.

Proofs

Quickly prove the identity theorem for holomorphic functions, i.e. suppose

  • $U$ is a domain
  • $f _ 1, f _ 2 : U \to \mathbb C$ are holomorphic functions
  • $S = \{z \in U \mid f _ 1(z) = f _ 2(z)\}$
  • $S$ has a limit point in $U$

Then

  • $f _ 1 = f _ 2$.

Overall idea: Decompose $S$ into isolated zeroes $I$ and non-isolated zeroes $N$. Then you can show that the non-isolated zeroes $N$ is actually both open (since it is the interior of some set) and closed (it is equal to its own closure). But then since $N$ is a subset of a connected set $U$, either $N$ is empty or $N$ is all of $U$. In the former case, this is a contradiction, since then $S$ would contain no limit points. In the latter case, we have the required result.

Let $g = f _ 1 - f _ 2$ so that $S = g^{-1}(\{0\})$. We want to show $S = U$.

Since $g$ is holomorphic on $U$, for any $z _ 0 \in S$, either $z _ 0$ is $I$solated or lies in an open ball contained in $S$ (by the theorem characterising the zeroes of holomorphic functions), call these points $N$on-isolated.

It then follows

\[S = I \cup N\]

where

  • $I = \{z \in S \mid z \text{ isolated}\}$
  • $N = \{z \in S \mid z \text{ is not isolated}\,\}$.

Note that $N = \text{int}(S)$, so $N$ is an open set.

Since $g$ is continuous, $g^{-1}(\{0\})$ must be closed in $U$ (since the preimage of a closed set by a continuous function must also be closed).

Thus $I \cup N$ is closed.

This implies $\text{Closure} _ U(N)$ (the closure of $N$ in $U$) must lie in $I \cup N$.

However, no limit point of $N$ can lie in $I$, so we must have $\text{Closure} _ U(N) = N$.

Thus $N$ is both open and closed in $U$, and since $U$ is connected, this implies either $N = \emptyset$ or $N = U$.

If $N = \emptyset$, all zeroes of $g$ are isolated, so $S$ would have no limit points.

Hence $N = U$, so $S = U$ as required.



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