KU Math 800: Lecture 1/23/2026
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Holomorphic Functions
Last time, we saw that holomorphic functions satisfy CR equations. A natural question is if the converse holds. The answer is no.
[example=%counter%] Let $f(x + iy) = \sqrt{|x||y|}$. Then $u = \sqrt{|x||y|}$ and $v = 0$ satisfy the CR equations but it turns out that $f$ fails to be homolomorphic. [/example]
The assumption that we do need is that $u$ and $v$ are $C^1$.
[theorem] Let $\Omega \subseteq \CC$ be open and suppose $f = u + iv:\Omega \to \CC$ with $u, v \in C^1$ and $u$ and $v$ satisfy the CR equations. Then $f$ is holomorphic. [/theorem]
Power Series
The motivation for studying power series is that they provide lots of examples of holomorphic functions.
[example=%counter%]
\[e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.\][/example]
[definition=%counter%] Let $f = \sum a_n z^n \in \CC[[z]]$. Then $f$ converges absolutely if $\sum |a_n||z|^n < \infty$. [/definition]
[theorem=%counter%] Given $\sum a_n z^n \in \CC[[z]]$, then there exists $R \in [0, \infty]$ such that:
- $|z| < R$ implies the series converges absolutely; and
- $|z| > R$ implies the series diverges.
Furthermore, $R^{-1} = \limsup_{n} \sqrt[n]{|a_n|}$. [/theorem]
[example=%counter%] The radius of convergence of $e^z = \sum \frac{z^n}{n!}$ is $R = \infty$. [/example]
[proof] By the theorem above, we have
\[\frac{1}{R} = \limsup_{n\to\infty} \left(\frac{1}{n!}\right)^{1/n} = \limsup_{n\to\infty} e^{\frac{1}{n}\ln\frac{1}{n!}}\]By continuity, it suffices to show that $\frac{1}{n}\ln \frac{1}{n!} \to -\infty$, which is clear. So $\frac{1}{R} = 0$ which implies $R = \infty$. [/proof]
[example=%counter%] The radius of convergence of $\sum z^n$ is $R = 1$. [/example]
[proof] By the theorem above, we clearly have $\frac{1}{R} = 1$ so $R = 1$. [/proof]
[definition=%counter%] Let $f:\Omega \to \CC$ be holomorphic. We say that $f$ is analytic at $z_0$ provided that
\[f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n\]for $z$ in a neighborhood of $z_0$. [/definition]
[corollary=%counter%] Analytic implies holomorphic. [/corollary]
Integration along Curves
[definition=%counter%] A parametrized curve is a function $z:[a, b] \to \CC$. It is smooth if $z$ is $C^1$ and $z’(t) \neq 0$ for all $t \in [a, b]$. It is closed if $z(a) = z(b)$. It is simple if it has no self-intersections except possibly at $a$ and $b$.
By convention, when we say that $\gamma$ is a smooth curve, we mean that it is a piecewise smooth curve in $\CC$. [/definition]
[definition=%counter%] Let $f$ be continuous on the smooth curve $\gamma$. Then
\[\int_{\gamma} f(z) \dd{z} \coloneqq \int_{a}^{b} f(z(t)) z'(t) \dd{t}.\][/definition]
[exercise=%counter%] Check this is independent of parametrization. [/exercise]
[proposition=%counter%]
- $\int_\gamma (\alpha f + \beta g) = \alpha \int_\gamma f + \beta \int_\gamma g$
- If $\gamma^{-}$ is $\gamma$ with reversed parametrization, then $\int_{\gamma^-} f= -\int_\gamma f$
- $|\int_\gamma f| \le \sup_{z\in\gamma} |f(z)| \cdot \operatorname{length}(\gamma)$. [/proposition]
[remark=%counter%] As a reminder,
\[\operatorname{length}(\gamma) \coloneqq \int_{a}^{b} |z'(t)| \dd{t}.\][/remark]