KU Math 800: Lecture 2/2/2026

Back to main page

On this page:

[corollary=%counter% (Corollary 4.3 in Stein-Shakarchi)] Let $f$ be holomorphic in an open set that contains the closure of a disc $D$ centered at $z_0$ and of radius $R$. Then

\[|f^{(n)}(z_0)| \le \frac{n!}{R^n} \cdot \sup_{z \in \partial D} |f(z)|.\]

[/corollary]

THe proof of this fact is immediate by Cauchy Integral Formulae and the usual integral-length-sup inequality.

[corollary=%counter% (Liouville’s theorem, Corollary 4.5 in Stein-Shakarchi)] If $f$ is entire and bounded, then $f$ is constant. [/corollary]

[proof] Notice that it suffices to show that $f’ = 0$ since $\CC$ is connected. By the Cauchy inequality,

\[|f'(z_0)| \le \frac{B}{R}\]

where $B$ is a bound for $f$ and $R$ is the radius of a disc $D$ centered at $z_0$. Since this holds for all $R > 0$, let $R \to \infty$ and the claim follows. [/proof]

[remark=%counter%] A counterexample for the analogous statement for $\RR$ is $f = \sin x$. [/remark]

[corollary=%counter% (Fundamental theorem of algebra, Corollary 4.6 in Stein-Shakarchi)] Every nonconstant $P \in \CC[z]$ has a root in $\CC$. [/corollary]

[proof] For the sake of contradiction, suppose that $P$ has no roots. Then $\frac{1}{P(z)}$ is an entire function. We also claim that it is bounded. Suppose that $P$ is degree $n$. Then we write

\[\frac{P(z)}{z^n} = a_n + \left(\frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right)\]

for $z \neq 0$. Each term in the parentheses in goes to $0$ as $|z| \to \infty$. So there exists $R > 0$ so that if $c = |a_n|/2$, then

\[|P(z)| \ge c|z|^n\]

whenever $|z| > R$. So $P$ is bounded from below so $P$ is bounded from above for $|z| > R$. For the $|z| \le R$, then $P$ being continuous implies it is bounded on the compact set $|z| \le R$. Now apply Liouville to obtain that $P$ is constant, contradiction. [/proof]

[corollary=%counter% (Corollary 4.7 in Stein-Shakarchi)] Every $P \in \CC[z]$ fully factors over $\CC$ into linear factors. [/corollary]

[corollary=%counter% (Corollary 4.9 in Stein-Shakarchi)] If $f$ and $g$ are holomorphic in a region $\Omega$ and $f(z) = g(z)$ for all $z$ in some non-empty open subset of $\Omega$, then $f(z) = g(z)$ throughout $\Omega$. [/corollary]

[proof] Suffices to show that if $f$ is holomorphic in a region $\Omega$ and vanishes on a sequence of distinct points with a limit point in $\Omega$, then $f$ is identically $0$. And for that, it suffices to prove that

\[U = \mathrm{Int}\{z\in\Omega \mid f(z) = 0\}\]

is closed in $\Omega$ since $\Omega$ is connected. Let $z_0 \in \Omega$ such that

\[z_0 = \lim_{i \to \infty} z_i\]

and $f(z_i) = 0$. It suffices to show that $f \equiv 0$ in a small disc containing $z_0$. Choose disc $D$ centered at $z_0$ and $D \subseteq \Omega$. Since $f$ is holormophic, it is analytic so we can expand it into a power series expansion

\[f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n.\]

If $f$ is not identically zero, then there exists a smallest integer $m$ such that $a_m \neq 0$. But then we can write

\[f(z) = a_m(z - z_0)^m (1 + g(z - z_0))\]

where $g(z - z_0) \to 0$ as $z \to z_0$. By assumption we know that $f(z_i) = 0$ where $\lim_{i \to \infty} z_i = z_0$. So we have a contradiction because $f(z_i) \neq 0$ from representation above. So $f(z) =0$.
[/proof]

[remark=%counter%] Notice that this says nonzero holomorphic functions behave a lot like polynomials! This does NOT say holomorphic functions cannot have infinitely many zeros. For example, $f(z) = \sin z$. [/remark]

Possible midterm topics

  • Schwarz Reflection Principle
  • Morera’s Theorem
  • Runge’s approximation theorem.

Chapter 3 preview:

  • Meromorphic functions.
  • Complex log.
  • Residue theorem.