1 files changed, 32 insertions, 43 deletions
diff --git a/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.md b/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.md
index ccdd0ea..3cf81f7 100644
--- a/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.md
+++ b/pages/complex-analysis/one-complex-variable/cauchys-integral-formula.md
@@ -3,39 +3,33 @@ title: Cauchy's Integral Formula
 parent: One Complex Variable
 grand_parent: Complex Analysis
 nav_order: 3
-# cspell:words
 ---
 
 # {{ page.title }}
 
-{: .theorem-title }
-> {{ page.title }}
->
-> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
-> Let $\gamma$ be a contour in $G$ such that the interior of $\gamma$ is contained in $G$.
-> Then for any point $a$ in the interior of $\gamma$,
->
-> $$
-> f(a) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} \, dz.
-> $$
-> {: .katex-display .mb-0 }
+{% theorem * Cauchy's Integral Formula %}
+Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
+Let $\gamma$ be a contour in $G$ such that the interior of $\gamma$ is contained in $G$.
+Then for any point $a$ in the interior of $\gamma$,
+
+$$
+f(a) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} \, dz.
+$$
+{% endtheorem %}
 
 {% proof %}
 {% endproof %}
 
-{: .theorem-title }
-> {{ page.title }} (Generalization)
-> {: #cauchys-integral-formula-generalized }
->
-> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
-> Then the $n$th derivative $f^{(n)}$ exists for every $n \in \NN$.
-> If $\gamma$ is a contour in $G$ such that the interior of $\gamma$ is contained in $G$,
-> then for any point $a$ in the interior of $\gamma$,
->
-> $$
-> f^{(n)} (a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz.
-> $$
-> {: .katex-display .mb-0 }
+{% theorem * Cauchy's Integral Formula (Generalization) %}
+Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
+Then the $n$th derivative $f^{(n)}$ exists for every $n \in \NN$.
+If $\gamma$ is a contour in $G$ such that the interior of $\gamma$ is contained in $G$,
+then for any point $a$ in the interior of $\gamma$,
+
+$$
+f^{(n)} (a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz.
+$$
+{% endtheorem %}
 
 {% proof %}
 {% endproof %}
@@ -50,20 +44,17 @@ and is often used to compute the integral.
 
 ## Many Consequences
 
-{: .theorem-title }
-> Cauchy's Estimate 
-> {: #cauchys-estimate }
->
-> Let $f$ be holomorphic on an open set containing the disc with center $a$ and radius $r>0$.
-> Then
->
-> $$
-> \norm{f^{(n)}(a)} \le \frac{n!}{r^n} \sup_{\abs{z-a} = r} \norm{f(z)} \qquad \forall n \in \NN.
-> $$
-> {: .katex-display .mb-0 }
+{% theorem * Cauchy's Estimate %}
+Let $f$ be holomorphic on an open set containing the disc with center $a$ and radius $r>0$.
+Then
+
+$$
+\norm{f^{(n)}(a)} \le \frac{n!}{r^n} \sup_{\abs{z-a} = r} \norm{f(z)} \qquad \forall n \in \NN.
+$$
+{% endtheorem %}
 
 {% proof %}
-From [{{ page.title }}](#cauchys-integral-formula-generalized)
+From [{{ page.title }}](#cauchy-s-integral-formula-generalization)
 for the circular contour around $a$ with radius $r$ we obtain
 
 $$
@@ -82,16 +73,14 @@ and the right hand side of the inequality reduces to the desired expression.
 
 Recall that an *entire* function is a holomorphic function that is defined everywhere in the complex plane.
 
-{: .theorem-title }
-> Liouville's Theorem
-> {: #liouvilles-theorem }
->
-> Every bounded entire function is constant.
+{% theorem * Liouville's Theorem %}
+Every bounded entire function is constant.
+{% endtheorem %}
 
 {% proof %}
 Consider an entire function $f$ and assume that $\norm{f(z)} \le M$ for all $z \in \CC$ and some $M > 0$.
 Since $f$ is holomorphic on the whole plane, we may make
-[Cauchy's Estimate](#cauchys-estimate)
+[Cauchy's Estimate](#cauchy-s-estimate)
 for all disks centered at any point $a \in \CC$ and with any radius $r>0$.
 For the first derivative, we have $\norm{f'(a)} \le M/r$, which tends to $0$ for $r \to \infty$.
 Hence $f' = 0$ in the whole plane. This