From 1d700ab734aa9b6711eb31796beb25cb7659d8e0 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 20 Dec 2022 20:11:25 +0100
Subject: More improvements to knowledge base

---
 source/know/concept/holomorphic-function/index.md | 46 +++++++++++++----------
 1 file changed, 26 insertions(+), 20 deletions(-)

(limited to 'source/know/concept/holomorphic-function')

diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md
index cf252c0..976758b 100644
--- a/source/know/concept/holomorphic-function/index.md
+++ b/source/know/concept/holomorphic-function/index.md
@@ -9,13 +9,13 @@ 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.
+is called **holomorphic** or **analytic** if it is **complex differentiable**
+in the vicinity 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.
+they are very "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,
@@ -23,14 +23,17 @@ and for all directions of $$\Delta z$$:
 
 $$\begin{aligned}
     \boxed{
-        f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
+        f'(z)
+        = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
     }
 \end{aligned}$$
 
 We decompose $$f$$ into the real functions $$u$$ and $$v$$ of real variables $$x$$ and $$y$$:
 
 $$\begin{aligned}
-    f(z) = f(x + i y) = u(x, y) + i v(x, y)
+    f(z)
+    = f(x + i y)
+    = u(x, y) + i v(x, y)
 \end{aligned}$$
 
 Since we are free to choose the direction of $$\Delta z$$, we choose $$\Delta x$$ and $$\Delta y$$:
@@ -56,9 +59,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-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.
+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.
 
 
 
@@ -70,7 +73,8 @@ provided that $$f(z)$$ is holomorphic for all $$z$$ in the area enclosed by $$C$
 
 $$\begin{aligned}
     \boxed{
-        \oint_C f(z) \dd{z} = 0
+        \oint_C f(z) \dd{z}
+        = 0
     }
 \end{aligned}$$
 
@@ -86,34 +90,36 @@ $$\begin{aligned}
     &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
 \end{aligned}$$
 
-Using Green's theorem, we integrate over the area $$A$$ enclosed by $$C$$:
+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.
+Since $$f(z)$$ is holomorphic, $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations,
+such that the integrands disappear and the final result is zero.
 {% include proof/end.html id="proof-int-theorem" %}
 
 
-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 $$z_0$$:
+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 $$z_0$$:
 
 $$\begin{aligned}
     \boxed{
-        f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+        f(z_0)
+        = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
     }
 \end{aligned}$$
 
 
 {% include proof/start.html id="proof-int-formula" -%}
-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:
+Thanks to the integral theorem, we know that
+the shape and size of $$C$$ are 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}
-- 
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