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Single-variable holomorphic function

[definition] Let $U \subseteq \CC$ be open. A function $f:U \to \CC$ is called holomorphic if for any point $z_0 \in U$, there exists a ball $B_\eps(z_0) \subseteq U$ of radius $\eps > 0$ around $z_0$ such that $f$ on $B_\eps(z_0)$ can be written as a convergent power series, i.e.

[/definition]

Cauchy-Riemann equations

[theorem] Write $z = x + iy$ and $f:U \to \CC$ as $f(x, y) = u(x, y) + iv(x, y)$. Then $f$ is holomorphic if and only if $u$ and $v$ are $C^1$ and

on $U$. [/theorem]

Wirtinger operators

Recall that the Wirtinger operators are the differential operators

Then the Cauchy-Riemann equations are equivalent to $\pdv{f}{\overline{z}} = 0$.

Relationship to real Jacobian

Let $U \subseteq \CC = \RR^2$ be an open subset. Then we can regard the (real) differentiable map $f:U \to \CC$ as a map $U \to \RR^2$. Then its differential

can be represented in the standard coordinates $z = x + iy$ (at $z \in U$) and $w = r + is$ (at $f(z)$) as the real Jacobian

We can extend $\dd{f(z)}$ to a $\CC$-linear map by extension of scalars:

At this point, we choose different bases: namely $\pdv{z}$ and $\pdv{\overline z}$. Thus, the differential is represented by the matrix

Using the fact that $\pdv{\overline f}{\overline z} = \overline{\left(\pdv{f}{z}\right)}$, $f$ being holomorphic implies that above matrix is

Cauchy integral formula

[theorem] $f:U \to \CC$ is holomorphic if and only if $f$ is continuously differentiable and for any $B_\eps(z_0) \subseteq U$

[/theorem]

Other standard single-variable theorems

Maximum (modulus) principle

[theorem] Let $U \subseteq \CC$ be open and connected. If $f:U \to \CC$ is holomorphic and non-constant, then $|f|$ has no local maximum in $U$. If $U$ is bounded and $f$ can be extended to a continuous function $f:\overline{U} \to \CC$, then $|f|$ attains its maximum on $\partial U$. [/theorem]

Identity theorem

[theorem] If $f, g:U \to \CC$ are two holomorphic function son a connected open subset $U\subseteq \CC$ such that $f(z) = g(z)$ for all $z$ in a nonempty open subset $V \subseteq U$, then $f = g$ on $U$. [/theorem]

Riemann extension theorem

[theorem] Let $f:B_\eps(z_0) \setminus z_0 \to \CC$ be a bounded holomorphic function. Then $f$ can be extended to a holomorphic function $f:B_\eps(0) \to \CC$. [/theorem]

Riemann mapping theorem

[definition] Let $U$ and $V$ be open subsets of $\CC$. We say that $U$ and $V$ are biholomorphic if there is a bijective holomorphic map $f:U \to V$ with holomorphic inverse $f^{-1}:V \to U$. [/definition]

[theorem] Let $U \subseteq \CC$ be a simply-connected proper open subset. Then $U$ is biholomorphic to $B_1(0)$. [/theorem]

Liouville

[theorem=Liouville’s theorem] Every bounded holomorphic function $f:\CC \to \CC$ is constant. [/theorem]

[corollary] There is no biholomorphism of $\CC$ and $B_\eps(0)$ with $\eps < \infty$. [/corollary]

[proof] Any holomorphic map $f:\CC \to B_{\eps}(0) \subseteq \CC$ were such a map is bounded and Liouville’s theorem implies that $f$ must be constant. So certainly $f$ is not a bijection, let alone a biholomorphism. [/proof]

Definition (polydisc)

[definition] Let $\eps \coloneqq (\eps_1, \ldots, e_n)$. We define the polydisc $B_\eps(w)$ to be the set

That is, a polydisc is a product of $\CC$-discs. [/definition]

Definition 1.1.1 (holomorphic function)

[definition] Let $U \subseteq \CC^n$ be an open subset and let $f:U \to \CC$ be continuously differentiable. Then $f$ is holomorphic if the Cauchy-Riemann equations holds for all coordinates $z_j = x_i + iy_j$, i.e.

where $j = 1, \ldots, n$. [/definition]

Thus, by definition, a continuously differentiable function $f$ is holomorphic if the induced functions

are holomorphic for all fixed $z_1, \ldots, z_{i-1}, z_{i+1}, \ldots, z_n \in \CC$.

Wirtinger formulation

If we define

then the multivariable Cauchy-Riemann equations can be restated as simply just that

for all $j = 1, \ldots, n$. For simplicity, we will write $\overline{\partial} f = 0$.

Proposition 1.1.2 (multivariable Cauchy integral formula)

[proposition] Let $f:\overline{B_\eps(w)} \to \CC$ be a continuous function such that $f$ is holomorphic with respect to every single component $z_i$ in any point of $B_\eps(w)$. Then, for any $z \in B_\eps(w)$,

[/proposition]

[proof] Just apply the single-variable Cauchy integral formula multiple times. Turning the iterated integral into the multiple integral follows by Fubini’s theorem since the integrand is continuous on the boundary of $B_\eps(w)$. [/proof]

Osgood’s lemma

Notice that if $f$ is holomorphic with respect to every single coordinate, then $f$ is necessarily holomorphic itself by just simply doing the usual “differentiation under the integral” trick that we do in single-variable complex analysis (recall that this is how we derive the Cauchy integral formulae for derivatives).

[corollary] If $f:U \to \CC$ is continuous and holomorphic with respect to every coordinate, then $f$ is holomorphic. [/corollary]

Power series expansion

Just like in single-variable complex analysis, the Cauchy integral formulae give formulas for the coefficients of the power series representation of a holomorphic function. In particular,

where

Single-variable theorems that generalize

The maximum modulus principle, identity theorem, and Liouville’s theorem generalize to the multivariable case. The Riemann extension theorem holds but is not trivial to prove, the Riemann mapping theorem fails badly.

Lemma 1.1.3

The holomorphicity of a function of serveral complex variables is typically shown by rrealizing it as an integral of a function known to be holomorphic.

[lemma] Let $U \subseteq \CC^n$ be an open subset and let $V \subseteq \CC$ be an open neighborhood of the booundary of $B_\eps(0) \subseteq \CC$. Assume that $f:V \times U \to \CC$ is a holomorphic function. Then

is a holomorphic function on $U$. [/lemma]

Proposition 1.1.4 (Hartogs’ theorem)

[theorem] Let $\eps = (\eps_1, \ldots, \eps_n)$ and $\eps’ = (\eps_1’, \ldots, \eps_n’)$ are given such that for all $i$, one has $\eps_i’ < \eps_i$. If $n > 1$, then any holomorphic map $f:B_\eps(0) \setminus \overline{B_{\eps’}(0)} \to \CC$ can be uniquely extended to a holomorphic map $f: B_\eps(0) \to \CC$. [/theorem]