# 2025-01-08-15-02-42

## Holomorphic functions of one variable

:::{.remark}
For $U \subseteq \CC$ open, let $f: U\to \CC$ be of class $C^1$.
Then $df = \dd{f}{x} dx + \dd{f}{y} dy$.
Writing $dz = dx + idy$ and $d\bar{z} = dx - idy$, one has
\[
dx = {1\over 2}(\dz + \dzbar) \quad dy = {1\over 2i}(\dz - \dzbar)
\]
and
\[
df = {1\over 2}(\dd{f}{x} - i \dd{f}{y})\dz + {1\over 2}(\dd{f}{x} + i\dd{f}{y})\dzbar
= \dd{f}{z}\dz + \dd{f}{\bar z}\dzbar
.\]

We define $f$ to be holomorphic if $\dd{f}{\zbar}=0$.
:::

:::{.lemma}
$f$ is holomorphic on $U$ iff $df$ is $\CC\dash$linear at every point in $U$.
:::

:::{.proof}
For $u\in U$, write $df_u = \dd{f}{x}(u)dx + \dd{f}{y}(u)dy$ as a map $\RR^2\cong \CC\to \CC$.
Note that $\dd{}{y} = i \dd{}{x}$.
Moreover, $df_u$ is $\CC\dash$linear if $df_u(\dd{}{y}) = i df_u(\dd{}{x})$, or equivalently $\dd{f}{y} = i \dd{f}{x}$.
One can rearrange this to $\dd{f}{x} - i \dd{f}{y} = 0$.
:::

:::{.lemma}
$f: U\to \CC$ is holomorphic iff $f\dz$ is a closed 1-form.
:::

:::{.proof}
One computes
\[
0

=d(f\dz) 
= df\wedge \dz 
= (\dd{f}{z}\dz + \dd{f}{\zbar}\dzbar)\wedge \dz
= \dd{f}{\zbar} \dzbar \wedge \dz
,\]
where we note 
\[
\dzbar \wedge \dz 
= (dx-idy)\wedge (dx+idy)
= i dx\wedge dy - idy\wedge dx 
= 2i dx\wedge dy
\neq 0
.\]
So $\dd{f}{\zbar} \dzbar \wedge \dz = 0 \iff \dd{f}{\zbar} = 0$.
:::

:::{.remark}
Recall Stokes' theorem:
\[
\int_M d\alpha = \int_{\bd M} \alpha
.\]

If $\alpha$ is a closed 1-form on $U \subseteq \CC$, then $\int_{\gamma} \alpha$ is independent of deformations of $\gamma$ since
\[
\int_ \gamma \alpha - \int _{\gamma'} \alpha = \int_U d \alpha = 0
.\]
where $U$ is an annulus with boundary \( \gamma\union \gamma' \).
:::

:::{.theorem}
If $f: U\to \CC$ is holomorphic and $\gamma$ is a simple closed curve bounding a domain $D$ in $U$, then $\forall z_0\in D^\circ$, 
\[
f(z_0) = {1\over 2i\pi} \int_ \gamma {f(z) \over z-z_0}\dz
.\]
:::

:::{.proof}
$g(z) \da f(z)\over z-z_0$ is holomorphic on $U\smts{z_0}$, so $g(z)\dz$ is a closed 1-form.
Thus the path of integration can be deformed to a small loop around $z_0$, say one parameterized by
\[
z \da z_0 + \eps e^{i\theta} \implies \dz = \eps ie^{i \theta}d\theta
.\]
One can now compute 
\[
{1\over 2i\pi}\int_ \gamma {f(z) \over z-z_0}\dz
= {1\over 2i\pi}\int_{\theta = 0}^{2\pi} { f(z_0 + \eps e^{i \theta}) \over \eps e^{i \theta} } \eps i e^{i \theta}d\theta 
= {1\over 2\pi} \int_0^{2\pi} f(z_0 + \eps e^{i \theta}) d\theta 
\convergesto{\eps\to 0} f(z_0)
.\]
:::

:::{.corollary}
If $f: U\to \CC$ is holomorphic, then $f$ is $C^\infty$.
Moreover, $f$ is complex analytic, i.e. $\forall z_0\in U, \exists R > 0$ such that $f(z_0 + z) = \sum_{n\geq 0} {f^{(n)}(z_0) \over n! } z^n$ for all $\abs{z} < R$.
:::

:::{.proof}
Write 
\[
f(z_0 + z) 
= {1\over 2i\pi}\int_ \gamma {f( \zeta) \over \zeta - z_0 - z} d\zeta
= {1\over 2i\pi}\int_ \gamma {1\over \zeta - z_0} {1\over 1 - {z\over \zeta-z_0}} f(\zeta)d\zeta 
= {1\over 2i\pi}\int_ \gamma {1\over \zeta - z_0} \sum_{n\geq 0} {z^n \over (\zeta-z_0)^n} f(\zeta)d\zeta 
= \sum_{n\geq 0}{1\over 2i\pi}\qty{ \int_ \gamma {f(\zeta)\over (\zeta-z_0)^{n+1}} d\zeta }z^n
.\]

:::

## Many complex variables

:::{.remark}
For $U \subseteq \CC^n$ is open and $f: U\to \CC$, write $\CC^n \cong \RR^{2n}$ with coordinates $x_1,\cdots, x_n, y_1, \cdots, y_n$ and $z_i = x_i + i y_i$.
Then $f = f(x_1, y_1, \cdots, x_n, y_n) = f(z_1, \bar z_1, \cdots, z_n, \bar z_n)$ is a function of $z_i$ and $\bar z_i$ and thus $\dd{f}{z_i}, \dd{f}{\bar z_i}$ make sense.
We say $f$ is holomorphic if $\dd{f}{\bar z_i} = 0$ for all $i$.
:::

:::{.lemma}
$f$ is holomorphic iff $f dz_1\wedge \cdots \wedge dz_n$ is closed.
:::

:::{.theorem}
For $z = (z_1,\cdots, z_n)\in U \subseteq \CC^n$,
\[
f(z) = {1\over (2i\pi)^n} \int_{\prod_{i=1}^n \ts{ \abs{\zeta_i - z_i} = a_i}} f(\zeta) {d\zeta_1\over\zeta_1 - z_1 } \wedge \cdots \wedge {d\zeta_n \over \zeta_n - z_n}
.\]

:::

:::{.theorem}
Let $f: U\to \CC^n$ be holomorphic on an open set.
If $\abs{f}$ admits a  maximum at a point $u\in U$, then $f$ is constant in a neighborhood of $u$.
:::

:::{.theorem}
If $f$ is holomorphic on a hyperplane complement $U \smts{z \st z_1 = 0}$ and $f$ is locally bounded near $\ts{z\st z_1=0}$ then $f$ extends to a holomorphic function on $U$.
:::


:::{.theorem}
Let $n\geq 2$.
If $f$ is holomorphic on $U\smts{z\st z_1 = z_2 = 0}$, then $f$ extends to a holomorphic function on $U$.
:::

:::{.remark}
Note that in the $n=2$ case, this says that a holomorphic function on $\CC^2\smz$ extends to a holomorphic function on $\CC^2$.
:::

:::{.proof} 
Consider the $n=2$ case where $U$ contains the origin.
If $f$ were holomorphic, Cauchy's formula would hold:
\[
f(z) = {1\over (2i\pi)^2}\int_{\abs{\zeta_i}=r_i}f(\zeta) {d\zeta_1\over \zeta_1-z_1}\wedge {d\zeta_2\over \zeta_2-z_2}
.\]
Replace this with
\[

f(z) = {1\over (2i\pi)^2}\int_{\abs{\zeta_i-z_i}=\eps_i}
f(\zeta) {d\zeta_1\over \zeta_1-z_1}\wedge {d\zeta_2\over \zeta_2-z_2}
,\]
which amounts to replacing the curve around the origin with a curve around...

:::