KU Math 800: Lecture 2/9/2026

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[definition=%counter%] Let $\Omega \subseteq \CC$ be an open subset. We say that $f:\Omega \to \CC$ is meromorphic on $\Omega$ if there exists a sequence of points $z_0, z_1, \ldots$ that has no limit points in $\Omega$ and:

1. $f$ is holomorphic on $\Omega\setminus \{z_0, z_1, \ldots\}$
2. $f$ has poles at $z_0, z_1, \ldots$. [/definition]

[theorem=%counter% (Argument Principle)] Suppose that $f$ is meromorphic in an open set containing a circle $C$ and its interior. Assume that $f$ has no poles and never vanishes in $\CC$. Then

counting multiplicity. [/theorem]

[proof] Say $z_0$ is a zero of $f$ of multiplicity $n$. Then

locally near $z_0$ where $g(z_0) \neq 0$ and $g$ is holomorphic near $z_0$. So then

So then $f’/f$ has a simple pole at $z_0$ so then $\operatorname{res}_{z_0} f’/f = n$.

Similarly, if $z_0$ is a pole of multiplicitly $n$, then

which implies that $f’/f$ has a simple pole at $z_0$ with residue $-n$. Now just apply Cauchy’s Residue Theorem and the claim follows. [/proof]

[remark=%counter%] For some intuition, notice that naively we can see that

Unfortunately, $\log f$ is not well-defined but the derivative is (because any branch has the same derivative). [/remark]

[corollary=%counter% (Theorem 4.3, Rouche’s Theorem)] Suppose $f, g$ are holomorphic in an open set $\Omega \subseteq \CC$. In addition, assume that $C \subseteq \Omega$ and that $f, g$ are holomorphic on $C$ and its interior. If

for all $z \in C$, then the number of zeros of $f$ in $C$ is the number of zeros of $f+g$ in $C$. [/corollary]

[proof] Because $f, g$ have no poles, it suffices to prove

by Argument Principle. The idea here is that we will deform $f$ to $f + g$ via linear homotopy. Define $f_t$ by

Now let $n_t$ be the number of zeros of $f_t$ in $C$. It suffices to show that $n_t$ is constant in $t$. We claim that $f_t(z)$ has no zeros on $C$. Indeed, if it did, then

which is a contradiction. So by the argument principle, we have

To prove $n_t$ is constant, it suffices to prove that $n_t$ is continuous in $t$. Indeed, the image of $n_t$ (in the variable $t$) is connected. So $n_t$ being continuous implies that $n_t$ takes on a single point. But it is clear that $n_t$ is continuous, because

is continuous as a function on $[0, 1] \times \Omega$. [/proof]

[definition=%counter%] A map is open if it takes open sets to open sets. [/definition]

[theorem=%counter% (Theorem 4.4, Open Mapping Theorem)] If $f:\Omega \to \CC$ is holomorphic and non-constant, then $f$ is open. [/theorem]

[proof] Let $w_0 = f(z_0)$ for some $z_0$. To show $f$ is open, it suffices to show that all points near $w_0$ also belong to image of $f$. Say $w$ is a nearby point of $w_0$ and define $g(z) = f(z) - w$. Then

Let $F(z) = f(z) - w_0$ and $G(z) = (w_0 - w)$. Choose $\delta > 0$ such that

and $f(z) \neq w_0$ on $|z - z_0| = \delta$. Then we can choose $\eps > 0$ so that

on the circle $|z - z_0| = \delta$. Suppose $|w - w_0| < \eps$. We want to show there exists $z \in \Omega$ such that $f(z) = \omega$. That’s equivalent to showing that $g(z)$ has a zero in $\Omega$. Notice that $F(z)$ has a zero in $C$. Clearly $F$ has a zero in $C$. To apply Rouche’s Theorem, it suffices to show that

for all $z\in C$. But since we assume that $|w - w_0| < \eps$ so then

[/proof]

[theorem=%counter% (Theorem 4.5, Maximum Modulus Principle)] Let $f:\Omega \to \CC$ be holomorphic and non-constant with $\Omega$ open. Then $|f|$ cannot attain a maximum on $\Omega$. [/theorem]

[proof] Suppose that $|f|$ has a maximum at $z_0 \in \Omega$. By the Open Mapping Theorem, since $f$ is open we can find a disc $D$ centered at $z_0$ so that $f(D)$ is open. This contradicts $f(z_0)$ having maximum modulus (you can always pick a further away $f(z) \in f(D)$ from the origin due to $f(D)$ being open). [/proof]