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pages/Math416/Math416_L4.md

@@ -146,21 +146,50 @@ EOP
 
 #### Theorem of differentiability
 
-Let $f:G\to \mathbb{C}$ be holomorphic function on open set $G\subset \mathbb{C}$ and real differentiable. $f=u+iv$ where $u,v$ are real differentiable functions.
+Let $f:G\to \mathbb{C}$ be a function defined on an open set $G\subset \mathbb{C}$ that is both holomorphic and (real) differentiable, where $f=u+iv$ with $u,v$ real differentiable functions.
 
-Then, $f$ is conformal if and only if $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0,\forall \zeta_0\in G$.
+Then, $f$ is conformal at every point $\zeta_0\in G$ if and only if $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$.
 
 Proof:
 
-<!---TODO: check after lecture-->
+We prove the equivalence in two parts.
 
-Case 1: Suppose $f(\zeta)=a\zeta+b\overline{\zeta}$, Let $b=\frac{\partial f}{\partial \overline{z}}(\zeta)$. We need to prove $a+b\neq 0$. So we want $b=0$ and $a\neq 0$, other wise $f(\mathbb{R})=0$.
+($\implies$) Suppose that $f$ is conformal at $\zeta_0$. By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $\zeta_0$. In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $\zeta_0$, $Df(\zeta_0)$, acts as a similarity transformation. Any similarity in $\mathbb{R}^2$ can be written as a rotation combined with a scaling; in particular, its matrix representation has the form
+$$
+\begin{pmatrix}
+A & -B \\
+B & A
+\end{pmatrix},
+$$
+for some real numbers $A$ and $B$. This is exactly the matrix corresponding to multiplication by the complex number $a=A+iB$. Therefore, the Cauchy-Riemann equations must hold at $\zeta_0$, implying that $f$ is holomorphic at $\zeta_0$. Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(\zeta_0)=a\neq 0$.
 
-$f:\mathbb{R}\to \{(a+b)t\}$ is not conformal.
+($\impliedby$) Now suppose that $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$. Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $\zeta_0$ is
+$$
+f(\zeta_0+h)=f(\zeta_0)+f'(\zeta_0)h+o(|h|),
+$$
+for small $h\in\mathbb{C}$. Multiplication by the nonzero complex number $f'(\zeta_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=\zeta_0$, we have
+$$
+(f\circ\gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0),
+$$
+and the angle between any two tangent vectors at $\zeta_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $\zeta_0$.
 
-...
+For further illustration, consider the special case when $f$ is an affine map.
 
-Case 2: Immediate consequence of the lemma of conformal function.
+Case 1: Suppose 
+$$
+f(\zeta)=a\zeta+b\overline{\zeta}.
+$$
+The Wirtinger derivatives of $f$ are
+$$
+\frac{\partial f}{\partial \zeta}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{\zeta}}=b.
+$$
+For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{\zeta}}=b=0$. Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$. If $b\neq 0$, then the map mixes $\zeta$ and $\overline{\zeta}$, and one can check that the linearization maps the real axis $\mathbb{R}$ into the set $\{(a+b)t\}$, which does not uniformly scale and rotate all directions. Thus, $f$ fails to be conformal when $b\neq 0$.
+
+Case 2: For a general holomorphic function, the lemma of conformal functions shows that if 
+$$
+(f\circ \gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0)
+$$
+for any differentiable curve $\gamma$ through $\zeta_0$, then the effect of $f$ near $\zeta_0$ is exactly given by multiplication by $f'(\zeta_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $\zeta_0$.
 
 EOP