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+---
+title: "Holomorphic function"
+date: 2021-02-25
+categories:
+- Mathematics
+- Complex analysis
+layout: "concept"
+---
+
+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 $C$:
+
+$$\begin{aligned}
+    \boxed{
+        \oint_C f(z) \dd{z} = 0
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-int-theorem"/>
+<label for="proof-int-theorem">Proof</label>
+<div class="hidden">
+<label for="proof-int-theorem">Proof.</label>
+Just like before, we decompose $f(z)$ into its real and imaginary parts:
+
+$$\begin{aligned}
+    \oint_C f(z) \dd{z}
+    &= \oint_C (u + i v) \dd{(x + i y)}
+    = \oint_C (u + i v) \:(\dd{x} + i \dd{y})
+    \\
+    &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
+\end{aligned}$$
+
+Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
+
+$$\begin{aligned}
+    \oint_C f(z) \dd{z}
+    &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
+\end{aligned}$$
+
+Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+equations, such that the integrands disappear and the final result is zero.
+</div>
+</div>
+
+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 $z_0$:
+
+$$\begin{aligned}
+    \boxed{
+        f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-int-formula"/>
+<label for="proof-int-formula">Proof</label>
+<div class="hidden">
+<label for="proof-int-formula">Proof.</label>
+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 $z_0$:
+
+$$\begin{aligned}
+    \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+    &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
+    = f(z_0)
+\end{aligned}$$
+</div>
+</div>
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        f^{(n)}(z_0)
+        = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-diff-formula"/>
+<label for="proof-diff-formula">Proof</label>
+<div class="hidden">
+<label for="proof-diff-formula">Proof.</label>
+By definition, the first derivative $f'(z)$ of a
+holomorphic function exists and is:
+
+$$\begin{aligned}
+    f'(z_0)
+    = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
+\end{aligned}$$
+
+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 $n$.
+</div>
+</div>
+