Merge branch 'main' of https://github.com/Trance-0/NoteNextra

pages/CSE559A/CSE559A_L5.md

@@ -0,0 +1,222 @@
+# Lecture 5
+
+## Continue on linear interpolation
+
+- In linear interpolation, extreme values are at the boundary.
+- In bicubic interpolation, extreme values may be inside.
+
+`scipy.interpolate.RegularGridInterpolator`
+
+### Image transformations
+
+Image warping is a process of applying transformation $T$ to an image.
+
+Parametric (global) warping: $T(x,y)=(x',y')$
+
+Geometric transformation: $T(x,y)=(x',y')$ This applies to each pixel in the same way. (global)
+
+#### Translation
+
+$T(x,y)=(x+a,y+b)$
+
+matrix form:
+
+$$
+\begin{pmatrix}
+x'\\y'
+\end{pmatrix}
+=
+\begin{pmatrix}
+1&0\\0&1
+\end{pmatrix}
+\begin{pmatrix}
+x\\y
+\end{pmatrix}
++
+\begin{pmatrix}
+a\\b
+\end{pmatrix}
+$$
+
+#### Scaling
+
+$T(x,y)=(s_xx,s_yy)$ matrix form:
+
+$$
+\begin{pmatrix}
+x'\\y'
+\end{pmatrix}
+=
+\begin{pmatrix}
+s_x&0\\0&s_y
+\end{pmatrix}
+\begin{pmatrix}
+x\\y
+\end{pmatrix}
+$$
+
+#### Rotation
+
+$T(x,y)=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$
+
+matrix form:
+
+$$
+\begin{pmatrix}
+x'\\y'
+\end{pmatrix}
+=
+\begin{pmatrix}
+\cos\theta&-\sin\theta\\\sin\theta&\cos\theta
+\end{pmatrix}
+\begin{pmatrix}
+x\\y
+\end{pmatrix}
+$$
+
+To undo the rotation, we need to rotate the image by $-\theta$. This is equivalent to apply $R^T$ to the image.
+
+#### Affine transformation
+
+$T(x,y)=(a_1x+a_2y+a_3,b_1x+b_2y+b_3)$
+
+matrix form:
+
+$$
+\begin{pmatrix}
+x'\\y'
+\end{pmatrix}
+=
+\begin{pmatrix}
+a_1&a_2&a_3\\b_1&b_2&b_3
+\end{pmatrix}
+\begin{pmatrix}
+x\\y\\1
+\end{pmatrix}
+$$
+
+Taking all the transformations together.
+
+#### Projective homography
+
+$T(x,y)=(\frac{ax+by+c}{gx+hy+i},\frac{dx+ey+f}{gx+hy+i})$
+
+$$
+\begin{pmatrix}
+x'\\y'\\1
+\end{pmatrix}
+=
+\begin{pmatrix}
+a&b&c\\d&e&f\\g&h&i
+\end{pmatrix}
+\begin{pmatrix}
+x\\y\\1
+\end{pmatrix}
+$$
+
+### Image warping
+
+#### Forward warping
+
+Send each pixel to its new position and do the matching.
+
+- May cause gaps where the pixel is not mapped to any pixel.
+
+#### Inverse warping
+
+Send each new position to its original position and do the matching.
+
+- Some mapping may not be invertible.
+
+#### Which one is better?
+
+- Inverse warping is better because it usually more efficient, doesn't have a problem with holes.
+- However, it may not always be possible to find the inverse mapping.
+
+## Sampling and Aliasing
+
+### Naive sampling
+
+- Remove half of the rows and columns in the image.
+
+Example:
+
+When sampling a sine wave, the result may interpret as different wave.
+
+#### Nyquist-Shannon sampling theorem
+
+- A bandlimited signal can be uniquely determined by its samples if the sampling rate is greater than twice the maximum frequency of the signal.
+
+- If the sampling rate is less than twice the maximum frequency of the signal, the signal will be aliased.
+
+#### Anti-aliasing
+
+- Sample more frequently. (not always possible)
+- Get rid of all frequencies that are greater than half of the new sampling frequency.
+  - Use a low-pass filter to get rid of all frequencies that are greater than half of the new sampling frequency. (eg, Gaussian filter)
+
+```python
+import scipy.ndimage as ndimage
+def down_sample(height, width, image):
+    # Apply Gaussian blur to the image
+    im_blur = ndimage.gaussian_filter(image, sigma=1)
+    # Down sample the image by taking every second pixel
+    return im_blur[::2, ::2]
+```
+
+## Nonlinear filtering
+
+### Median filter
+
+Replace the value of a pixel with the median value of its neighbors.
+
+- Good for removing salt and pepper noise. (black and white dot noise)
+
+### Morphological operations
+
+Binary image: image with only 0 and 1.
+
+Let $B$ be a structuring element and $A$ be the original image (binary image).
+
+- Erosion: $A\ominus B = \{p\mid B_p\subseteq A\}$, this is the set of all points that are completely covered by $B$.
+- Dilation: $A\oplus B = \{p\mid B_p\cap A\neq\emptyset\}$, this is the set of all points that are at least partially covered by $B$.
+- Opening: $A\circ B = (A\ominus B)\oplus B$, this is the set of all points that are at least partially covered by $B$ after erosion.
+- Closing: $A\bullet B = (A\oplus B)\ominus B$, this is the set of all points that are completely covered by $B$ after dilation.
+
+Boundary extraction: use XOR operation on eroded image and original image.
+
+Connected component labeling: label the connected components in the image. _use prebuild function in scipy.ndimage_
+
+## Light,Camera/Eyes, and Color
+
+### Principles of grouping and Gestalt Laws
+
+- Proximity: objects that are close to each other are more likely to be grouped together.
+- Similarity: objects that are similar are more likely to be grouped together.
+- Closure: objects that form a closed path are more likely to be grouped together.
+- Continuity: objects that form a continuous path are more likely to be grouped together.
+
+### Light and surface interactions
+
+A photon's life choices:
+
+- Absorption
+- Diffuse reflection (nice to model) (lambertian surface)
+- Specular reflection (mirror-like) (perfect mirror)
+- Transparency
+- Refraction
+- Fluorescence (returns different color)
+- Subsurface scattering (candles)
+- Photosphorescence
+- Interreflection
+
+#### BRDF (Bidirectional Reflectance Distribution Function)
+
+$$
+\rho(\theta_i,\phi_i,\theta_o,\phi_o)
+$$
+
+- $\theta_i$ is the angle of incidence.
+- $\phi_i$ is the azimuthal angle of incidence.
+- $\theta_o$ is the angle of reflection.
+- $\phi_o$ is the azimuthal angle of reflection.

pages/CSE559A/CSE559A_L6.md

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+# Lecture 6
+
+## Continue on Light, eye/camera, and color
+
+### BRDF (Bidirectional Reflectance Distribution Function)
+
+$$
+\rho(\theta_i,\phi_i,\theta_o,\phi_o)
+$$
+
+#### Diffuse Reflection
+
+- Dull, matte surface like chalk or latex paint
+
+- Most often used in computer vision
+- Brightness _does_ depend on direction of illumination
+
+Diffuse reflection governed by Lambert's law: $I_d = k_d N\cdot L I_i$
+
+- $N$: surface normal
+- $L$: light direction
+- $I_i$: incident light intensity
+- $k_d$: albedo
+
+$$
+\rho(\theta_i,\phi_i,\theta_o,\phi_o)=k_d \cos\theta_i
+$$
+
+#### Photometric Stereo
+
+Suppose there are three light sources, $L_1, L_2, L_3$, and we have the following measurements:
+
+$$
+I_1 = k_d N\cdot L_1
+$$
+
+$$
+I_2 = k_d N\cdot L_2
+$$
+
+$$
+I_3 = k_d N\cdot L_3
+$$
+
+We can solve for $N$ by taking the dot product of $N$ and each light direction and then solving the system of equations.
+
+Will not do this in the lecture.
+
+#### Specular Reflection
+
+- Mirror-like surface
+
+$$
+I_e=\begin{cases}
+I_i & \text{if } V=R \\
+0 & \text{if } V\neq R
+\end{cases}
+$$
+
+- $V$: view direction
+- $R$: reflection direction
+- $\theta_i$: angle between the incident light and the surface normal
+
+Near-perfect mirror have a high light around $R$.
+
+common model:
+
+$$
+I_e=k_s (V\cdot R)^{n_s}I_i
+$$
+
+- $k_s$: specular reflection coefficient
+- $n_s$: shininess (imperfection of the surface)
+- $I_i$: incident light intensity
+
+#### Phong illumination model
+
+- Phong approximation of surface reflectance
+  - Assume reflectance is modeled by three compoents
+    - Diffuse reflection
+    - Specular reflection
+    - Ambient reflection
+
+$$
+I_e=k_a I_a + I_i \left[k_d (N\cdot L) + k_s (V\cdot R)^{n_s}\right]
+$$
+
+- $k_a$: ambient reflection coefficient
+- $I_a$: ambient light intensity
+- $k_d$: diffuse reflection coefficient
+- $k_s$: specular reflection coefficient
+- $n_s$: shininess
+- $I_i$: incident light intensity
+
+Many other models.
+
+#### Measuring BRDF
+
+Use Gonioreflectometer.
+
+- Device for measuring the reflectance of a surface as a function of the incident and reflected angles.
+- Can be used to measure the BRDF of a surface.
+
+BRDF dataset:
+
+- MERL dataset
+- CURET dataset
+
+### Camera/Eye
+
+#### DSLR Camera
+
+- Pinhole camera model
+- Lens
+- Aperture (the pinhole)
+- Sensor
+- ...
+
+#### Digital Camera block diagram
+
+![Digital Camera block diagram](https://static.notenextra.trance-0.com/images/CSE559A/DigitalCameraBlockDiagram.png)
+
+Scanning protocols:
+
+- Global shutter: all pixels are exposed at the same time
+- Interlaced: odd and even lines are exposed at different times
+- Rolling shutter: each line is exposed as it is read out
+

pages/CSE559A/_meta.js

@@ -7,4 +7,6 @@ export default {
     CSE559A_L2: "Computer Vision (Lecture 2)",
     CSE559A_L3: "Computer Vision (Lecture 3)",
     CSE559A_L4: "Computer Vision (Lecture 4)",
+    CSE559A_L5: "Computer Vision (Lecture 5)",
+    CSE559A_L6: "Computer Vision (Lecture 6)",
 }

pages/Math4111/Math4111_L24.md

@@ -133,6 +133,28 @@ By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\i
 
 EOP
 
+---
+
+Supplemental materials:
+
+_I found this section is not covered in the lecture but is used in later chapters._
+
+#### Definition 4.18
+
+Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. $f$ is **uniformly continuous** on $X$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x, y\in X$, $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.
+
+#### Theorem 4.19
+
+If $f$ is a continuous mapping of a compact metric space $X$ into a metric space $Y$, then $f$ is uniformly continuous on $X$.
+
+Proof:
+
+See the textbook.
+
+EOP
+
+---
+
 ### Continuity and connectedness
 
 > **Definition 2.45**: Let $X$ be a metric space. $A,B\subset X$ are **separated** if $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.

pages/Math4121/Math4121_L7.md

@@ -1 +1,96 @@
-# Lecture 7
+# Lecture 7
+
+## Continue on Chapter 6
+
+### Riemann integrable
+
+#### Theorem 6.6
+
+A function $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a, b]$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
+
+Proof:
+
+$\impliedby$
+
+For every $P$,
+
+$$
+L(f, P, \alpha) \leq \underline{\int}_a^b f d\alpha \leq \overline{\int}_a^b f d\alpha \leq U(f, P, \alpha)
+$$
+
+So if $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$, then for every $\epsilon > 0$, there exists a partition $P$ such that
+
+$$
+0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha \leq U(f, P, \alpha) - L(f, P, \alpha) < \epsilon
+$$
+
+Thus $0 \leq \overline{\int}_a^b f d\alpha - \underline{\int}_a^b f d\alpha < \epsilon,\forall \epsilon > 0$.
+
+Then, $\overline{\int}_a^b f d\alpha = \underline{\int}_a^b f d\alpha$.
+
+So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+$\implies$
+
+If $f\in \mathscr{R}(\alpha)$ on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+Then by the definition of Riemann integrable, $\sup_{P} L(f, P, \alpha) =\int^b_a f d\alpha = \inf_{P} U(f, P, \alpha)$.
+
+Given any $\epsilon > 0$, by definition of infimum and supremum, there exists a partition $P_1,P_2$ such that
+
+$$
+\int^b_a f d\alpha - \frac{\epsilon}{2} < L(f, P_1, \alpha) \leq \sup_{P} L(f, P, \alpha) = \inf_{P} U(f, P, \alpha) < \int^b_a f d\alpha + \frac{\epsilon}{2}
+$$
+
+Taking $P = P_1 \cup P_2$, by [Theorem 6.4](https://notenextra.trance-0.com/Math4121/Math4121_L6#theorem-64) we have
+
+$$
+U(f, P, \alpha) - L(f, P, \alpha) \leq \left( \int^b_a f d\alpha + \frac{\epsilon}{2} \right) - \left( \int^b_a f d\alpha - \frac{\epsilon}{2} \right) = \epsilon
+$$
+
+So $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+EOP
+
+#### Theorem 6.8
+
+If $f$ is continuous on $[a, b]$, then $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+Proof:
+
+> Main idea:
+>
+> $$U(f, P, \alpha) - L(f, P, \alpha) = \sum_{i=1}^n \left( M_i - m_i \right) \Delta \alpha_i$$
+>
+> If we can make $M_i - m_i$ small enough, then $U(f, P, \alpha) - L(f, P, \alpha)$ can be made arbitrarily small.
+>
+> Since $M_i=\sup_{x\in [t_{i-1}, t_i]} f(x)$ and $m_i=\inf_{x\in [t_{i-1}, t_i]} f(x)$, we can make $M_i - m_i$ small enough by making the partition $P$ sufficiently fine.
+
+Suppose we can find a partition $P$ such that $M_i - m_i < \eta$. Then $U(f, P, \alpha) - L(f, P, \alpha) \leq\eta\sum_{i=1}^n \Delta \alpha_i = \eta (\alpha(b)-\alpha(a))$.
+
+> Let $\epsilon >0$ and choose $\eta = \frac{\epsilon}{\alpha(b)-\alpha(a)}$. Then there exists a partition $P$ such that $U(f, P, \alpha) - L(f, P, \alpha) < \epsilon$.
+
+Since $f$ is continuous on $[a, b]$ (a compact set), then $f$ is uniformly continuous on $[a, b]$. [Theorem 4.19](https://notenextra.trance-0.com/Math4111/Math4111_L24#theorem-419)
+
+> If $f$ is continuous on $x$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.
+>
+> If $f$ is continuous on $[a, b]$, then $f$ is continuous at $x,\forall x\in [a, b]$.
+
+So, there exists a $\delta > 0$ such that for all $x, t\in [a, b]$ with $|x-t| < \delta$, we have $|f(x)-f(t)| < \eta$.
+
+Let $P=\{x_0, x_1, \cdots, x_n\}$ be a partition of $[a, b]$ such that $\Delta x_i < \delta$ for all $i$.
+
+So, $\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| < \eta$ for all $i$.
+
+$$
+\begin{aligned}
+\sup_{x,t\in [x_{i-1}, x_i]} |f(x)-f(t)| &= \sup_{x,t\in [x_{i-1}, x_i]} f(x)-f(t) \\
+&= \sup_{x\in [x_{i-1}, x_i]} f(x)-\sup_{t\in [x_{i-1}, x_i]} -f(t) \\
+&=\sup_{x\in [x_{i-1}, x_i]} f(x)-\inf_{t\in [x_{i-1}, x_i]} f(t) \\
+&= M_i - m_i
+\end{aligned}
+$$
+
+So, $f$ is Riemann integrable with respect to $\alpha$ on $[a, b]$.
+
+EOP

pages/Math416/Math416_L5.md

@@ -0,0 +1,225 @@
+# Lecture 5
+
+## Review
+
+Let $f$ be a complex function. that maps $\mathbb{R}^2$ to $\mathbb{R}^2$. $f(x+iy)=u(x,y)+iv(x,y)$.
+
+$Df(x+iy)=\begin{pmatrix}
+\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
+\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
+\end{pmatrix}=\begin{pmatrix}
+\alpha & \beta\\
+\sigma & \delta
+\end{pmatrix}$
+
+So 
+
+$$\begin{aligned}
+\frac{\partial f}{\partial \zeta}&=\frac{1}{2}\left(u_x+v_y\right)-i\frac{1}{2}\left(v_x+u_y\right)\\
+&=\frac{1}{2}\left(\alpha+\delta\right)-i\frac{1}{2}\left(\beta-\sigma\right)\\
+\end{aligned}$$
+
+$$
+\begin{aligned}
+\frac{\partial f}{\partial \overline{\zeta}}&=\frac{1}{2}\left(u_x+v_y\right)+i\frac{1}{2}\left(v_x+u_y\right)\\
+&=\frac{1}{2}\left(\alpha-\delta\right)+i\frac{1}{2}\left(\beta+\sigma\right)\\
+\end{aligned}
+$$
+
+When $f$ is conformal, $Df(x+iy)=\begin{pmatrix}
+\alpha & \beta\\
+-\beta & \alpha
+\end{pmatrix}$.
+
+So $\frac{\partial f}{\partial \zeta}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a$
+
+$\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}(\alpha-\alpha)+i\frac{1}{2}(\beta-\beta)=0$
+
+> Less pain to represent a complex function using four real numbers.
+
+## Chapter 3: Linear fractional Transformations
+
+Let $a,b,c,d$ be complex numbers. such that $ad-bc\neq 0$.
+
+The linear fractional transformation is defined as
+
+$$
+\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}
+$$
+
+If we let $\psi(\zeta)=\frac{e\zeta-f}{-g\zeta+h}$ also be a linear fractional transformation, then $\phi\circ\psi$ is also a linear fractional transformation.
+
+New coefficients can be solved by
+
+$$
+\begin{pmatrix}
+a & b\\
+c & d
+\end{pmatrix}
+\begin{pmatrix}
+e & f\\
+g & h
+\end{pmatrix}
+=
+\begin{pmatrix}
+k&l\\
+m&n
+\end{pmatrix}
+$$
+
+So $\phi\circ\psi(\zeta)=\frac{k\zeta+l}{m\zeta+n}$
+
+### Complex projective space
+
+$\mathbb{R}P^1$ is the set of lines through the origin in $\mathbb{R}^2$.
+
+We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{R}^2\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{R}\setminus\{0\}$ such that $(a,b)=t(c,d)$.
+
+$R\mathbb{P}^1=S^1\setminus\{\pm x\}\cong S^1$
+
+Equivalently,
+
+$\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
+
+We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{C}\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{C}\setminus\{0\}$ such that $(a,b)=(tc,td)$.
+
+So, $\forall \zeta\in\mathbb{C}\setminus\{0\}$:
+
+If $a\neq 0$, then $(a,b)\sim(1,\frac{b}{a})$.
+
+If $a=0$, then $(0,b)\sim(0,-b)$.
+
+So, $\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
+
+### Linear fractional transformations
+
+Let $M=\begin{pmatrix}
+a & b\\
+c & d
+\end{pmatrix}$ be a $2\times 2$ matrix with complex entries. That maps $\mathbb{C}^2$ to $\mathbb{C}^2$.
+
+Suppose $M$ is non-singular. Then $ad-bc\neq 0$.
+
+If $M\begin{pmatrix}
+\zeta_1\\
+\zeta_2
+\end{pmatrix}=\begin{pmatrix}
+\omega_1\\
+\omega_2
+\end{pmatrix}$, then $M\begin{pmatrix}
+t\zeta_1\\
+t\zeta_2
+\end{pmatrix}=\begin{pmatrix}
+t\omega_1\\
+t\omega_2
+\end{pmatrix}$.
+
+So, $M$ induces a map $\phi_M:\mathbb{C}P^1\to\mathbb{C}P^1$ defined by $M\begin{pmatrix}
+\zeta\\
+1
+\end{pmatrix}=\begin{pmatrix}
+\frac{a\zeta+b}{c\zeta+d}\\
+1
+\end{pmatrix}$.
+
+$\phi_M(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+If we let $M_2=\begin{pmatrix}
+e &f\\
+g &h
+\end{pmatrix}$, where $ad-bc\neq 0$ and $eh-fg\neq 0$, then $\phi_{M_2}(\zeta)=\frac{e\zeta+f}{g\zeta+h}$.
+
+So, $M_2M_1=\begin{pmatrix}
+a&b\\
+c&d
+\end{pmatrix}\begin{pmatrix}
+e&f\\
+g&h
+\end{pmatrix}=\begin{pmatrix}
+\zeta\\
+1
+\end{pmatrix}$.
+
+This also gives $\begin{pmatrix}
+k\zeta+l\\
+m\zeta+n
+\end{pmatrix}\sim\begin{pmatrix}
+\frac{k\zeta+l}{m\zeta+n}\\
+1
+\end{pmatrix}$.
+
+So, if $ab-cd\neq 0$, then $\exists M^{-1}$ such that $M_2M_1=I$.
+
+So non-constant linear fractional transformations form a group under composition.
+
+When do two matrices gives the $t_0$ same linear fractional transformation?
+
+$M_2^{-1}M_1=\alpha I$
+
+We defined $GL(2,\mathbb{C})$ to be the group of general linear transformations of order 2 over $\mathbb{C}$.
+
+This is equivalent to the group of invertible $2\times 2$ matrices over $\mathbb{C}$ under matrix multiplication.
+
+Let $F$ be the function that maps $M$ to $\phi_M$.
+
+$F:GL(2,\mathbb{C})\to\text{Homeo}(\mathbb{C}P^1)$
+
+So the kernel of $F$ is the set of matrices that represent the identity transformation. $\ker F=\left\{\alpha I\right\},\alpha\in\mathbb{C}\setminus\{0\}$.
+
+#### Corollary of conformality
+
+If $\phi$ is a non-constant linear fractional transformation, then $\phi$ is conformal.
+
+Proof:
+
+Know that $\phi_0\circ\phi(\zeta)=\zeta$,
+
+Then $\phi(\zeta)=\phi_0^{-1}\circ\phi\circ\phi_0(\zeta)$.
+
+So $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+$\phi:\mathbb{C}\cup\{\infty\}\to\mathbb{C}\cup\{\infty\}$ which gives $\phi(\infty)=\frac{a}{c}$ and $\phi(-\frac{d}{c})=\infty$.
+
+So, $\phi$ is conformal.
+
+EOP
+
+#### Proposition 3.4 of Fixed points
+
+Any non-constant linear fractional transformation except the identity transformation has 1 or 2 fixed points.
+
+Proof:
+
+Let $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+Case 1: $c=0$
+
+Then $\infty$ is a fixed point.
+
+Case 2: $c\neq 0$
+
+Then $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+The solution of $\phi(\zeta)=\zeta$ is $c\zeta^2+(d-a)\zeta-b=0$.
+
+Such solutions are $\zeta=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
+
+So, $\phi$ has 1 or 2 fixed points.
+
+EOP
+
+#### Proposition 3.5 of triple transitivity
+
+If $\zeta_1,\zeta_2,\zeta_3\in\mathbb{C}P^1$ are distinct, then there exists a non-constant linear fractional transformation $\phi$ such that $\phi(\zeta_1)=\zeta_2$ and $\phi(\zeta_3)=\infty$.
+
+Proof as homework.
+
+#### Theorem 3.8 Preservation of clircles
+
+We defined clircle to be a circle or a line.
+
+If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
+
+Proof:
+
+Continue on next lecture.

pages/Math416/Math416_L6.md

@@ -0,0 +1,219 @@
+# Lecture 6
+
+## Review
+
+### Linear Fractional Transformations
+
+Transformations of the form $f(z)=\frac{az+b}{cz+d}$,$a,b,c,d\in\mathbb{C}$ and $ad-bc\neq 0$ are called linear fractional transformations.
+
+#### Theorem 3.8 Preservation of clircles
+
+We defined clircle to be a circle or a line.
+
+The circle equation is:
+
+Let $\zeta=u+iv$ be the center of the circle, $r$ be the radius of the circle.
+
+$$
+circle=\{z\in\mathbb{C}:|\zeta-c|=r\}
+$$
+
+This is:
+
+$$
+|\zeta|^2-c\overline{\zeta}-\overline{c}\zeta+|c|^2-r^2=0
+$$
+
+If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
+
+We claim that a map is circle preserving if and only if for some $\alpha,\beta,\gamma,\delta\in\mathbb{R}$.
+
+$$
+\alpha|\zeta|^2+\beta Re(\zeta)+\gamma Im(\zeta)+\delta=0
+$$
+
+when $\alpha=0$, it is a line.
+
+when $\alpha\neq 0$, it is a circle.
+
+Proof:
+
+Let $w=u+iv=\frac{1}{\zeta}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
+
+Then the original equation becomes:
+
+$$
+\alpha\left(\frac{u}{u^2+v^2}\right)^2+\beta\left(\frac{u}{u^2+v^2}\right)+\gamma\left(-\frac{v}{u^2+v^2}\right)+\delta=0
+$$
+
+Which is in the form of circle equation.
+
+EOP
+
+## Chapter 4 Elements of functions
+
+> $e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}$
+
+So, following the definition of $e^\zeta$, we have:
+
+$$
+\begin{aligned}
+e^{x+iy}&=e^xe^{iy} \\
+&=e^x\left(\sum_{n=0}^{\infty}\frac{(iy)^n}{n!}\right) \\
+&=e^x\left(\sum_{n=0}^{\infty}\frac{(-1)^ny^n}{n!}\right) \\
+&=e^x(\cos y+i\sin y)
+\end{aligned}
+$$
+
+### $e^\zeta$
+
+The exponential of $e^\zeta=x+iy$ is defined as:
+
+$$
+e^\zeta=exp(\zeta)=e^x(\cos y+i\sin y)
+$$
+
+So,
+
+$$
+|e^\zeta|=|e^x||\cos y+i\sin y|=e^x
+$$
+
+#### Theorem 4.3 $e^\zeta$ is holomorphic
+
+$e^\zeta$ is holomorphic on $\mathbb{C}$.
+
+Proof:
+
+$$
+\begin{aligned}
+\frac{\partial}{\partial\zeta}e^\zeta&=\frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{i}{\partial y}\right)e^x(\cos y+i\sin y) \\
+&=\frac{1}{2}e^x(\cos y+i\sin y)+ie^x(-\sin y+i\cos y) \\
+&=0
+\end{aligned}
+$$
+
+EOP
+
+#### Theorem 4.4 $e^\zeta$ is periodic
+
+$e^\zeta$ is periodic with period $2\pi i$.
+
+Proof:
+
+$$
+e^{\zeta+2\pi i}=e^\zeta e^{2\pi i}=e^\zeta\cdot 1=e^\zeta
+$$
+
+EOP
+
+#### Theorem 4.5 $e^\zeta$ as a map
+
+$e^\zeta$ is a map from $\mathbb{C}$ to $\mathbb{C}$ with period $2\pi i$.
+
+$$
+e^{\pi i}+1=0
+$$
+
+This is a map from cartesian coordinates to polar coordinates, where $e^x$ is the radius and $y$ is the angle.
+
+This map attains every value in $\mathbb{C}\setminus\{0\}$.
+
+#### Definition 4.6-8 $\cos\zeta$ and $\sin\zeta$
+
+$$
+\cos\zeta=\frac{1}{2}(e^{i\zeta}+e^{-i\zeta})
+$$
+
+$$
+\sin\zeta=\frac{1}{2i}(e^{i\zeta}-e^{-i\zeta})
+$$
+
+$$
+\cosh\zeta=\frac{1}{2}(e^\zeta+e^{-\zeta})
+$$
+
+$$
+\sinh\zeta=\frac{1}{2}(e^\zeta-e^{-\zeta})
+$$
+
+From this definition, we can see that $\cos\zeta$ and $\sin\zeta$ are no longer bounded.
+
+And this definition is still compatible with the previous definition of $\cos$ and $\sin$ when $\zeta$ is real.
+
+Moreover,
+
+$$
+\cosh(i\zeta)=\cos\zeta
+$$
+
+$$
+\sinh(i\zeta)=i\sin\zeta
+$$
+
+### Logarithm
+
+#### Definition 4.9 Logarithm
+
+A logarithm of $a$ is any $b$ such that $e^b=a$.
+
+If $a=0$, then no logarithm exists.
+
+If $a\neq 0$, then there exists infinitely many logarithms of $a$.
+
+Let $a=re^{i\theta}$, $b=x+iy$ be a logarithm of $a$.
+
+Then,
+
+$$
+e^{x+iy}=re^{i\theta}
+$$
+
+Since logarithm is not unique, we can always add $2k\pi i$ to the angle.
+
+If $y\in(-\pi,\pi]$, then $\log a=b$ means $e^b=a$ and $Im(b)\in(-\pi,\pi]$.
+
+If $a=re^{i\theta}$, then $\log a=\log r+i(\theta_0+2k\pi)$.
+
+#### Definition 4.10
+
+Let $G$ be an open connected subset of $\mathbb{C}\setminus\{0\}$.
+
+A branch of $\arg(\zeta)$ in $G$ is a continuous function $\alpha$, such that $\alpha(\zeta)$ is a value of $\arg(\zeta)$.
+
+A branch of $\log(\zeta)$ in $G$ is a continuous function $\beta$, such that $e^{\beta(\zeta)}=\zeta$.
+
+Note: $G$ has a branch of $\arg(\zeta)$ if and only if it has a branch of $\log(\zeta)$.
+
+If $G=\mathbb{C}\setminus\{0\}$, then  not branch of $\arg(\zeta)$ exists.
+
+Suppose $\alpha_1$ and $\alpha_2$ are two branches of $\arg(\zeta)$ in $G$.
+
+Then,
+
+$$
+\alpha_1(\zeta)-\alpha_2(\zeta)=2k\pi i
+$$
+
+for some $k\in\mathbb{Z}$.
+
+#### Theorem 4.11
+
+$\log(\zeta)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
+
+Proof:
+
+Method 1: Use polar coordinates. (See in homework)
+
+Method 2: Use the fact that $\log(\zeta)$ is the inverse of $e^\zeta$.
+
+Suppose $h=s+it$, $e^h=e^s(\cos t+i\sin t)$, $e^h-1=e^s(\cos t-1)+i\sin t$. So
+
+$$
+\begin{aligned}
+\frac{e^h-1}{h}&=\frac{(s+it)e^s(\cos t-1)+i\sin t}{s^2+t^2} \\
+&=\frac{e^s(\cos t-1)}{s^2+t^2}+i\frac{\sin t}{s^2+t^2}
+\end{aligned}
+$$
+
+Continue next time.

pages/Math416/_meta.js

@@ -7,4 +7,6 @@ export default {
     Math416_L2: "Complex Variables (Lecture 2)",
     Math416_L3: "Complex Variables (Lecture 3)",
     Math416_L4: "Complex Variables (Lecture 4)",
+    Math416_L5: "Complex Variables (Lecture 5)",
+    Math416_L6: "Complex Variables (Lecture 6)",
 }