From c705ac1d7dc74709835a8c48fae4a7dd70dc5c49 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 25 Feb 2021 16:14:20 +0100
Subject: Expand knowledge base

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+---
+title: "Holomorphic function"
+firstLetter: "H"
+publishDate: 2021-02-25
+categories:
+- Mathematics
+
+date: 2021-02-25T14:40:45+01:00
+draft: false
+markup: pandoc
+---
+
+# Holomorphic function
+
+In complex analysis, a complex function $f(z)$ of a complex variable $z$
+is called **holomorphic** or **analytic** if it is complex differentiable in the
+neighbourhood of every point of its domain.
+This is a very strong condition.
+
+As a result, holomorphic functions are infinitely differentiable and
+equal their Taylor expansion at every point. In physicists' terms,
+they are extremely "well-behaved" throughout their domain.
+
+More formally, a given function $f(z)$ is holomorphic in a certain region
+if the following limit exists for all $z$ in that region,
+and for all directions of $\Delta z$:
+
+$$\begin{aligned}
+    \boxed{
+        f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
+    }
+\end{aligned}$$
+
+We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:
+
+$$\begin{aligned}
+    f(z) = f(x + i y) = u(x, y) + i v(x, y)
+\end{aligned}$$
+
+Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:
+
+$$\begin{aligned}
+    f'(z)
+    &= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x}
+    = \pdv{u}{x} + i \pdv{v}{x}
+    \\
+    &= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y}
+    = \pdv{v}{y} - i \pdv{u}{y}
+\end{aligned}$$
+
+For $f(z)$ to be holomorphic, these two results must be equivalent.
+Because $u$ and $v$ are real by definition,
+we thus arrive at the **Cauchy-Riemann equations**:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{u}{x} = \pdv{v}{y}
+        \qquad
+        \pdv{v}{x} = - \pdv{u}{y}
+    }
+\end{aligned}$$
+
+Therefore, a given function $f(z)$ is holomorphic if and only if its real
+and imaginary parts satisfy these equations. This gives an idea of how
+strict the criteria are to qualify as holomorphic.
+
+
+## Integration formulas
+
+Holomorphic functions satisfy **Cauchy's integral theorem**, which states
+that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero,
+provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:
+
+$$\begin{aligned}
+    \boxed{
+        \oint_C f(z) \dd{z} = 0
+    }
+\end{aligned}$$
+
+*__Proof__*.
+*Just like before, we decompose $f(z)$ into its real and imaginary parts:*
+
+$$\begin{aligned}
+    \oint_C f(z) \:dz
+    &= \oint_C (u + i v) \dd{(x + i y)}
+    = \oint_C (u + i v) \:(\dd{x} + i \dd{y})
+    \\
+    &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
+\end{aligned}$$
+
+*Using Green's theorem, we integrate over the area $A$ enclosed by $C$:*
+
+$$\begin{aligned}
+    \oint_C f(z) \:dz
+    &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
+\end{aligned}$$
+
+*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+equations, such that the integrands disappear and the final result is zero.*
+*__Q.E.D.__*
+
+An interesting consequence is **Cauchy's integral formula**, which
+states that the value of $f(z)$ at an arbitrary point $z_0$ is
+determined by its values on an arbitrary contour $C$ around $z_0$:
+
+$$\begin{aligned}
+    \boxed{
+        f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+    }
+\end{aligned}$$
+
+*__Proof__*.
+*Thanks to the integral theorem, we know that the shape and size
+of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$,
+such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then
+we integrate by substitution:*
+
+$$\begin{aligned}
+    \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+    &= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta}
+    = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+\end{aligned}$$
+
+*We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:*
+
+$$\begin{aligned}
+    \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+    &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
+    = f(z_0)
+\end{aligned}$$
+
+*__Q.E.D.__*
+
+Similarly, **Cauchy's differentiation formula**,
+or **Cauchy's integral formula for derivatives**
+gives all derivatives of a holomorphic function as follows,
+and also guarantees their existence:
+
+$$\begin{aligned}
+    \boxed{
+        f^{(n)}(z_0)
+        = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z}
+    }
+\end{aligned}$$
+
+*__Proof__*.
+*By definition, the first derivative $f'(z)$ of a
+holomorphic function $f(z)$ exists and is given by:*
+
+$$\begin{aligned}
+    f'(z_0)
+    = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
+\end{aligned}$$
+
+*We evaluate the numerator using Cauchy's integral theorem as follows:*
+
+$$\begin{aligned}
+    f'(z_0)
+    &= \lim_{z \to z_0} \frac{1}{z - z_0}
+    \bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg)
+    \\
+    &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
+    \oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta}
+    \\
+    &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
+    \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta}
+\end{aligned}$$
+
+*This contour integral converges uniformly, so we may apply the limit on the inside:*
+
+$$\begin{aligned}
+    f'(z_0)
+    &= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta}
+    = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
+\end{aligned}$$
+
+*Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
+this proof works inductively for all higher orders $n$.*
+*__Q.E.D.__*
+
+
+## Residue theorem
+
+A function $f(z)$ is **meromorphic** if it is holomorphic except in
+a finite number of **simple poles**, which are points $z_p$ where
+$f(z_p)$ diverges, but where the product $(z - z_p) f(z)$ is non-zero and
+still holomorphic close to $z_p$.
+
+The **residue** $R_p$ of a simple pole $z_p$ is defined as follows, and
+represents the rate at which $f(z)$ diverges close to $z_p$:
+
+$$\begin{aligned}
+    \boxed{
+        R_p = \lim_{z \to z_p} (z - z_p) f(z)
+    }
+\end{aligned}$$
+
+**Cauchy's residue theorem** generalizes Cauchy's integral theorem
+to meromorphic functions, and states that the integral of a contour $C$,
+depends on the simple poles $p$ it encloses:
+
+$$\begin{aligned}
+    \boxed{
+        \oint_C f(z) \dd{z} = i 2 \pi \sum_{p} R_p
+    }
+\end{aligned}$$
+
+*__Proof__*. *From the definition of a meromorphic function,
+we know that we can decompose $f(z)$ as follows,
+where $h(z)$ is holomorphic and $p$ are all its poles:*
+
+$$\begin{aligned}
+    f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p}
+\end{aligned}$$
+
+*We integrate this over a contour $C$ which contains all poles, and apply
+both Cauchy's integral theorem and Cauchy's integral formula to get:*
+
+$$\begin{aligned}
+    \oint_C f(z) \dd{z}
+    &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
+    = \sum_{p} R_p \: 2 \pi i
+\end{aligned}$$
+
+*__Q.E.D.__*
+
+This theorem might not seem very useful,
+but in fact, thanks to some clever mathematical magic,
+it allows us to evaluate many integrals along the real axis,
+most notably [Fourier transforms](/know/concept/fourier-transform/).
+It can also be used to derive the Kramers-Kronig relations.
+
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