update notations and fix typos

pages/Math416/Math416_L4.md

@@ -14,9 +14,9 @@ $$
 
 ### Angle between two curves
 
-Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-The angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
+The angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
 
 ### Cauchy-Riemann equations
 
@@ -28,13 +28,13 @@ $$
 
 ### Theorem of conformality
 
-Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=z_0$ for some $t_0\in I_1\cap I_2$.
 
-If $f'(\zeta_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the same as the angle between the vectors $f'(\zeta_0)\gamma_1'(t_0)$ and $f'(\zeta_0)\gamma_2'(t_0)$.
+If $f'(z_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $z_0$ is the same as the angle between the vectors $f'(z_0)\gamma_1'(t_0)$ and $f'(z_0)\gamma_2'(t_0)$.
 
 ### Lemma of function of a curve and angle
 
-If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=\zeta_0$ for some $t_0\in I$.
+If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=z_0$ for some $t_0\in I$.
 
 Then,
 
@@ -60,12 +60,14 @@ $$
 > 
 > $f$ is differentiable if and only if $f(z+h)=f(z)+f'(z)h+\frac{1}{2}h^2f''(z)+o(h^3)$ as $h\to 0$. (By Taylor expansion)
 
-Since $f$ is holomorphic at $\gamma(t_0)=\zeta_0$, we have
+Since $f$ is holomorphic at $\gamma(t_0)=z_0$, we have
 
 $$
-f(\zeta_0)=f(\zeta_0)+(\zeta-\zeta_0)f'(\zeta_0)+o(\zeta-\zeta_0)
+f(z_0)=f(z_0)+(z-z_0)f'(z_0)+o(z-z_0)
 $$
 
+> This result comes from Taylor Expansion of the derivative of the function around the point $z_0$
+
 and
 
 $$
@@ -85,7 +87,7 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Definition 2.12 (Conformal function)
 
@@ -93,7 +95,7 @@ A function $f:G\to \mathbb{C}$ is called conformal if it preserves the angle bet
 
 #### Theorem 2.13 (Conformal function)
 
-If $f:G\to \mathbb{C}$ is conformal at $\zeta_0\in G$, then $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$.
+If $f:G\to \mathbb{C}$ is conformal at $z_0\in G$, then $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
 
 Example:
 
@@ -105,20 +107,20 @@ is not conformal at $z=0$ because $f'(0)=0$.
 
 #### Lemma of conformal function
 
-Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial \zeta}(\zeta_0)$, $b=\frac{\partial f}{\partial \overline{\zeta}}(\zeta_0)$.
+Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial z}(z_0)$, $b=\frac{\partial f}{\partial \overline{z}}(z_0)$.
 
-Let $\gamma(t_0)=\zeta_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
+Let $\gamma(t_0)=z_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
 
 Proof:
 
 $f=u+iv$, $u,v$ are real differentiable.
 
 $$
-a=\frac{\partial f}{\partial \zeta}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
+a=\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
 $$
 
 $$
-b=\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
+b=\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
 $$
 
 $$
@@ -131,7 +133,7 @@ $$
 
 $$
 \begin{aligned}
-(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial \zeta}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{\zeta}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
+(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial z}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{z}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
 &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\alpha}{dt}+i\frac{d\beta}{dt}\right)\\
 &+\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\beta}{dt}-i\frac{d\alpha}{dt}\right) \\
 &=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\alpha}{dt}-\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\beta}{dt}\right]\\
@@ -142,56 +144,56 @@ $$
 \end{aligned}
 $$
 
-EOP
+QED
 
 #### Theorem of differentiability
 
 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 at every point $\zeta_0\in G$ if and only if $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0$.
+Then, $f$ is conformal at every point $z_0\in G$ if and only if $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
 
 Proof:
 
 We prove the equivalence in two parts.
 
-($\implies$) 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
+($\implies$) Suppose that $f$ is conformal at $z_0$. By definition, conformality means that $f$ preserves angles (including their orientation) between any two intersecting curves through $z_0$. In the language of real analysis, this requires that the (real) derivative (Jacobian) of $f$ at $z_0$, $Df(z_0)$, 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 $f'(\zeta_0)=a\neq 0$.
+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 $z_0$, implying that $f$ is holomorphic at $z_0$. Moreover, because the transformation is nondegenerate (preserving angles implies nonzero scaling), we must have $f'(z_0)=a\neq 0$.
 
-($\impliedby$) 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
+($\impliedby$) Now suppose that $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$. Then by the definition of the complex derivative, the first-order (linear) approximation of $f$ near $z_0$ is
 $$
-f(\zeta_0+h)=f(\zeta_0)+f'(\zeta_0)h+o(|h|),
+f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|),
 $$
-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
+for small $h\in\mathbb{C}$. Multiplication by the nonzero complex number $f'(z_0)$ is exactly a rotation and scaling (i.e., a similarity transformation). Therefore, for any smooth curve $\gamma(t)$ with $\gamma(t_0)=z_0$, we have
 $$
-(f\circ\gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0),
+(f\circ\gamma)'(t_0)=f'(z_0)\gamma'(t_0),
 $$
-and the angle between any two tangent vectors at $\zeta_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $\zeta_0$.
+and the angle between any two tangent vectors at $z_0$ is preserved (up to the fixed rotation). Hence, $f$ is conformal at $z_0$.
 
 For further illustration, consider the special case when $f$ is an affine map.
 
 Case 1: Suppose 
 $$
-f(\zeta)=a\zeta+b\overline{\zeta}.
+f(z)=az+b\overline{z}.
 $$
 The Wirtinger derivatives of $f$ are
 $$
-\frac{\partial f}{\partial \zeta}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{\zeta}}=b.
+\frac{\partial f}{\partial z}=a \quad \text{and} \quad \frac{\partial f}{\partial \overline{z}}=b.
 $$
-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 $b\neq 0$.
+For $f$ to be holomorphic, we require $\frac{\partial f}{\partial \overline{z}}=b=0$. Moreover, to have a nondegenerate (angle-preserving) map, we must have $a\neq 0$. If $b\neq 0$, then the map mixes $z$ and $\overline{z}$, 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 
 $$
-(f\circ \gamma)'(t_0)=f'(\zeta_0)\gamma'(t_0)
+(f\circ \gamma)'(t_0)=f'(z_0)\gamma'(t_0)
 $$
-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 $\zeta_0$.
+for any differentiable curve $\gamma$ through $z_0$, then the effect of $f$ near $z_0$ is exactly given by multiplication by $f'(z_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $z_0$.
 
-EOP
+QED
 
 ### Harmonic function
 
@@ -227,7 +229,7 @@ $$
 \Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.
 $$
 
-EOP
+QED
 
 If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$.